Here's my solution, hope it's correct - it's very easy to make mistakes in combinatorial geometry. Consider the convex hull of the points. For one side $ A_{1}A_{2}$ (where these two points are the endpoints of that edge), consider the point $ A_{3}$ that maximizes the area of $ A_{1}A_{2}A_{3}$. Now consider the point $ D$ that completes the parallelogram of these 3 points. Since $ D$ must lie on, or inside the convex hull, it is clear that $ DA_{3}$ is parallel to $ A_{1}A_{2}$- so for each side in the convex hull polygon, there is a side parallel to it. For any edge $ A_{1}A_{2}$ consider the parallel edge $ B_{1}B_{2}$. Suppose they are not of equal length, Wlog $ A_{1}A_{2}$ is longer. Consider the point that completes the parallelogram $ A_{1}A_{2}B_{1}$ - this point must lie outside the segment $ B_{1}B_{2}$, which is a contradiction. Thus for any side, it must have a side parallel to it of equal length. Suppose on edge $ A_{1}A_{2}$, there exists another point lying on that segment, $ X$. Consider the edge parallel to $ A_{1}A_{2}$, $ B_{1}B_{2}$. Any point that completes the parallelogram $ B_{1}B_{2}X$ must lie outside the convex hull, our outside the segment $ A_{1}A_{2}$, contradiction. So no edge in the convex hull polygon contains 3, or more, vertices. Label the vertices $ A_{1}A_{2} ... A_{n}$, in cyclic order. Let $ A_{k}A_{k+1}$ be the parallel edge to $ A_{1}A_{2}$. Suppose $ A_{k+1} A_{k+2}$ is not parallel to $ A_{2}A_{3}$. Suppose instead, $ A_{t}A_{t+1}$ is parallel to $ A_{k+1}A_{k+2}$, with $ t \ge 3$. Consider the edge parallel to $ A_{2}A_{3}$. By drawing a diagram, we can see such an edge cannot be part of the convex hull - contradiction. Thus, we can group the vertices into pairs $ (A_{i}, B_{i})$, so that $ A_{i}A_{i+1}$ is parallel to $ B_{i}B_{i+1}$. Consider such a grouping. We can prove that the midpoint of $ A_{i}B_{i}$ is independent $ i$: Let $ O$ be the 'center' of parellelogram $ A_{1}A_{2}B_{1}B_{2}$. Since $ OA_{2} = OB_{2}$ and $ A_{2}A_{3}$ is parallel to, and equal in length, with, $ B_{2}B_{3}$. Thus $ O$ is also the 'centre' of parellelogram $ A_{2}A_{3}B_{2}B_{3}$. Proceed in the same fashion around the convex hull polygon. Now suppose the convex hull has $ >4$ points. Consider 3 points $ A_{1}A_{3}B_{2}$. It's easy to see any point completing the parellelogram of these $ 3$ points lies outside the convex hull, a contradiction. Thus the convex hull has $ 2$ points, in which case there are no other points in the set, (since these 2 points constitute the convex hull); OR there are $ 4$ points in the convex hull, constituting a parallelogram. If there is a point $ T$ inside this convex hull, consider the 3 points $ T, A_{1}, A_{2}$ - the point completing parallelogram must lie outside the convex hull, a contradiction. So we have our second solution set - $ 4$ points constituting a parellelogram. But we are not done yet - so far we have assumed not all the points are collinear. So we have the case when our points all lie on a line. Let the endpoints be $ A_{1}$ and $ B_{1}$. Let $ O$ be the midpoint of $ A_{1}B_{1}$. For any point $ X$ on the line, by considering the 3 points, $ X, A_{1}, B_{1}$, it's clear that the reflection of $ X$ about $ A_{1}B_{1}$ also lies in our set. Let $ A_{i}$ be the points left of $ O$, and $ B_{i}$ right of $ O$, so that $ A_{i}O = B_{i}O$. Suppose that $ O$ does not lie in our set - if it does, the following method with slight changes will work. Consider points $ A_{1}A_{i}B_{2}$ and $ A_{1}B_{i}B_{2}$, for $ i \ge 3$. Since $ A_{i}X > B_{1}B_{2}$, for each $ X=A_{3}, A_{4} ... B_{4}, B_{3}$. Let $ O'$ be the midpoint of $ A_{1}B_{2}$. For each point in the set $ A_{3} ... B_{3}$, the reflection of that point about $ O'$ also lies in the set. So we can pair up the points. It's clear that if say, $ Y$ lies to the left of $ X$, the reflection, $ Y'$ of $ Y$, lies to the right of the reflection, $ X'$, of $ X$. $ B_{3}$ can't be reflected to $ A_{3}$ or any point to the right of it, since $ A_{1}A_{3} > B{2}B{3}$; thus $ B_{3}$ is reflected to $ A_{2}$. Let $ x_{i} = A_{i}A_{i+1}$. The fact that $ B_{3}B_{2}=A_{1}A_{2}$, gives us $ x_{1}=x_{2}$. Next, $ B_{4}$ cannot go to $ A_{4}$ or any point to the right of it, since $ B_{2}B_{4} < A_{1}A_{4}$, thus $ B_{4}$ goes to $ A_{3}$, giving us $ x_{2}=x_{3}$. Continuing in this process, we have $ B_{k+1}$ goes to $ A_{k}$ for $ k=2,3...n-1$, and also $ A_{n}$ goes to itself. The equalities that arise from these facts, give us $ x_{1}=x_{2} = ... =x_{n-1}=A_{n}B_{n}$, so we have proven that all $ 2n$ points are equally spaced. If $ O$ is also part of the set, then we can prove in the same method that all $ 2n+1$ points are equally spaced. Thus we have a new solution set: any set $ n$ equally spaced points, for any integer $ n$. However, in this proof (of the part when points are collinear), we have assumed that there are at least $ 5$ points - if there are $ 4$ collinear points, they don't have to be equally spaced, as long as $ A_{1}A_{2}=A_{3}A_{4}$ where the points are $ A_{i}$ going from left to right. In conclusion solution sets: a parallelogram, a degenerate parellogram that is a line, and any set of equally spaced points.

VERY clever solution due to Ubemaya: We claim all sets S are 4 points that are vertices of a parallelogram, and sets with less than 3 points. Suppose there was a set with more than 4 points that would satisfy the conditions. Take the triangle of maximal area, and let it be ABC. A quadrilateral ABCD contains 4 triangles congruent to ABC, so it must be the convex hull, as a point outside ABCD would result in a triangle of greater area than ABC. There must be some 5th point E inside ABCD because there are more than 4 points. There must be another point F such that AEBF is a parallelogram. However F cannot be inside ABCD, which is the convex hull, so we have a contradiction, and so there cannot be a set with more than 4 points that satisfies the property.

The above post is too clever! here's a different, less slick way, similar to epitomy01. Let $|S| = n$. Let the convex hull of all $n$ points be $C$. Take two adjacent, distinct points on $C$, say $A$ and $B$. Let the point on $C$ with the maximal perpendicular distance to $\overline{AB}$ (where this means the whole line, not just the line segment) be $P$, and let this perpendicular distance be $d$. Let the slope of $\overline{AB} = \theta$, and the line with slope $\theta$ passing through $C$ be $l$. Observe that $l$ either passes through no points in $S$, or they pass through points in $S$ that have a perpendicular distance of $d$ to $\overline{AB}$, ie points in $S$ on $l$. This is because, if there's a point with perpendicular distance to $\overline{AB}$ greater than $d$, then it lies outside the convex hull by definition of $P$. To further rigorize, you can take a parallel line to $l$, say $l'$ with perpendicular distance to $\overline{AB}$ greater than $d$, then all the points on $l'$ are outside $C$, contradiction. Suppose $P$ does not lie on $\overline{AB}$, and that there exist no points in $S$ on $l$ besides $P$. Observe that for any three distinct points $M, N, O$ in plane, the only points with $M, N, O$ as vertices of a parallelogram are $M$ reflected over the midpoint of $NO$, $N$ reflected by the midpoint of $OM$, and $O$ reflected over midpoint of $MN$. Then if $P$ is reflected over $AB$ it lies outside $C$, and same for $B$ over $AP$, and $A$ over $BP$. But this means is a contradiction! Therefore, at least one more point lies on $l$. Case 1: no $3$ points collinear Let's change our notation. Let the first $m$ points in counterclockwise order be $A_1, A_2, \cdots, A_m$ and the next $p$ points in counterclockwise order be $B_1, B_2, \cdots, B_p$. By our work above, we have that $C$ is a convex polygon with $A_jA_{j+1} \parallel B_jB_{j+1}$ for all $j \le \min(m, n)$. WLOG, $\min{m, p} = p$. The definition of a parallelogram yields the midpoint of $A_iB_i$, say $O$, is fixed for all $i \le n$. Suppose for the sake of contradiction that $p \ge 3$. If $O$ lies outside $\triangle A_1A_3B_2$, this contradicts the counterclockwise ordering of the $A$s then $B$s, so $O$ lies inside $\triangle A_1A_3B_2$. Then $B_2O = A_2O$, so $B_2$ reflected over $A_1A_3$ lies outside $C$. Similarly, we have $A_1$ reflected over $A_3B_2$ and $A_3$ reflected over $A_1B_2$ lies outside $C$. Both of these cases are contradictions, so $p \le 2$. Hence $n \le 4$. The solution here is a parallelogram with $4$ points, any $2$ points, any $1$ point, or the empty set. Case 2: there exist $3$ collinear points. Let the points be $X_1, X_2, X_3$ on line $l_1$. If there exists no other points on $l_1$, then it is not hard to see that the solution set is an arbitrary number of equally spaced points on $l$. Else, there must exist a point not on $l_1$. Then there exists a point $Y_2$ such that $X_1X_2Y_1Y_2$ is a parallelogram. Let $\overline{Y_1Y_2} = l_2$. There must exist a similar point $Y_3$ on $l_2$. Suppose there are $k$ points on $l$. By a straightforward argument, then there exists $k$ points on $l_2$. Continue in this fashion, so we have $n$ lines $l_1, \cdots, l_n$ with $k$ points on each of those lines. Suppose that there are only $kn$ points in $S$. But then we can easily see that there is one more point in $S$ by the problem condition, so there are an infinite number of points in $S$, contradiction. Thus there are no solutions in this case. Therefore the solutions are an arbitrary number of equally spaced, distinct, collinear points, a parallelogram with $4$ points, any $2$ points, any $1$ point, or the empty set and we are done.

Solution with yunseo. Suppose $S$ has $n+1$ points for $n\ge 3$ and not all the points are collinear. Then, we have that no three $A$, $B$, $C$ are collinear, since there is no fourth $D$ to make a parallelogram. Thus, no three points are collinear. We now essentially reduce to and solve the one dimensional version of this problem. Clearly we may pick a line $\ell$ such that the projections of the points in $S$ onto $\ell$ are all distinct (there are only $\binom{n+1}{2}$ bad directions for $\ell$). Imposing a coordinate system on $\ell$, we may assume that the projections are \[S=\{0=x_0<x_1<x_2<\cdots<x_{n-1}<x_n\}.\]Suppose the point corresponding to $x_i$ is $A_i$. For any $1\le i\ne j\le n$, we have some $k\ne 0,i,j$ such that $A_0A_iA_jA_k$ is a parallelogram in some order. Thus, we have some $k\ne 0,i,j$ such that \[x_k\in\{x_i+x_j,x_i-x_j,x_j-x_i\}.\]This is actually enough to show that $n\le 3$. Note that $x_n+x_i>x_n$ and $x_i-x_n<0$, so we have $x_n-x_i\in S$. Thus, \[x_n-x_1>x_n-x_2>\cdots>x_n-x_{n-1}\]are all in $S$. Since $x_n-x_1>x_0=0$ and $x_n-x_1<x_n$, these must be \[x_{n-1}>x_{n-2}>\cdots>x_1,\]so we have $\boxed{x_i+x_{n-i}=x_n}$ for all $1\le i\le n-1$. We now do a similar more careful analysis with $x_{n-1}$. Note that $x_{n-1}+x_i>x_{n-1}+x_1=x_n$ for $i\ge 2$, so again, we have \[x_{n-1}-x_2>x_{n-1}-x_3>\cdots>x_{n-1}-x_{n-2}\]all in $S$. Note that $x_{n-1}-x_2<x_n-x_2=x_{n-2}$, so we have that these are exactly \[x_{n-3}>x_{n-4}>\cdots>x_1,\]so $\boxed{x_i+x_{n-1-i}=x_{n-1}}$ for all $1\le i\le n-2$. Subtracting the two boxed equations, we get \[x_{n-i}-x_{n-1-i}=x_n-x_{n-1}\]for all $1\le i\le n-2$ (assuming $n\ge 4$). This implies that $x_i$ is an arithmetic sequence, so we may write $x_i=di$ for some $d$. If $n$ is even, then we have $0,dn/2,dn\in S$. We have $dn+dn/2>dn$, so we have $dn-dn/2\in S$. But we want $x_k\in\{x_i+x_j,x_i-x_j,x_j-x_i\}$ for $k\ne i,j$, so this is not possible. Similarly, if $n\ge 5$ is odd, then $0,d(n-1)/2,d(n-1)$ are in $S$, and a similar argument gives a contradiction. Thus, we have $n\le 3$. So we have at most $4$ points, so the only allowed configurations are a parallelogram, and any configuration of at most $2$ points. We also have the configuration where all the points are collinear, but this doesn't work for three or more points. So the answer is any set with at most two points, along with a parallelogram.

We claim that $S$ can have atmost 4 points. The construction is easy. For the proof first , pick the Triangle $ABC$ which has maximal area. Now consider $D$ such that $ABCD$ is a parallelogram. Note that if there is a point $P$ outside $ABCD$ ,then atleast one of the areas of the triangles $ABP$ , $BCP$ ,$CDP$ ,$DAP$ must be larger than the area of $ABC$ (Contradiction!) .Similarly if P lies inside $ABCD$ ,then consider $P_1$ , $P_2$ , $P_3$ ,$P_4$ such that $APBP_1$ ,$BPCP_2$ , $CPDP_3$ ,$DPAP_4$ . It is clear that atleast one of the $P_i$ lie outside $ABCD$ ,and we arrive at a contradiction

One Question: Why are we allowed to include the cases with less than or equal to 2 points? You literally can't even choose 3 points from them so shouldn't they technically be invalid?