Lemma: If $A\in S$ and $O\neq A$ then a closed disc with radius $OA$ belongs to $S$. Proof: By the conditions of the problem we have that the circle $w$ with diameter $OA$ lies in $S$. Then take any point $B$ in the interior of this circle and draw a line $l$ perpendicular to $OB$ at $B$. Take $C\in l\cap w$ (note that the latter in non-empty) and draw the circle with diameter $OC$ ($C\in S$). Note that $B$ lies on this circle. Hence $B\in S$. Q.E.D. Now comes the proof of the problem. First of all let's consider the case when there is a ray $z$ passing through $O$ such that $\forall x\in \mathbb R^+\ \ \exists X\in S\colon X\in z,\ OX>x$. Draw a line $y\bot z : y\cap z=O$ and denote the half plane whose boundary is $y$ and in which $z$ lies by $U$. Note that every point that lies in $U$ but does not lie on $y$ is in $S$ (using the choice of $z$ we prove it in the same fasion as we have proven lemma). Now note that every point that lies on $y$ is in $S$ (take a point in $U$ sufficiently close to $y$ and sufficiently far from $O$, draw a circle with diameter whose one endpoint is the chosen point and the other is $O$ and use lemma). Now in the same manner (taking $y$ instead $z$) we prove that every point in each of the two half-planes in which $z$ and $-z$ divides the plane belongs to $S$. And finaly we prove that $-z$ lies in $S$ (again analogously). Hence $S$ coincides with the plane. That was the nasty part of the solution. Now comes the other one with a briliant idea. In this case that is left such ray $z$ as aforementioned does not exist and consequently $\exists m\in \mathbb{R^+}\colon \forall A\in S\ \ OA\le m$. And let $m$ be as minimal as it can be. In other words $m=\sup \{ |OA| \colon A\in S \}$. Now perform inversion centered at $O$ with radius $m$. Denote $X'$ an inversive image of $X$ and $S'$ an inversive image of $S$. Note that the conditions of the problem imply that for every $A'\in S'$ the line $l'\bot OA'\colon A'\in l'$ lies in $S'$ and our lemma implies that all half-plane which has $l'$ as the boundary and does not contain $O$ lies in $S'$. Now choose $l'$ sufficiently close to an open disc centered at $O$ with radius $m$. Fix $\epsilon >0$ and set $l_0=l'$. By $A_i$ we will denote a point on $l_i$ such that $OA_i\bot l_i$. Now choose $B_0\in l_0 \colon B_0A_0=\epsilon$ and let $l_1$ pass through $B_0$ perpendicularly to $OB_0$. $l_{n+1}$ is obtained from $l_n$ in the same fashion. We just have to choose such $B_n$ that $\angle A_{n-1}B_nA_{n+1}$ is obtuse. Note that in such a way we step by step rotate around the circle of inversion by a fixed a angle. And note that every $l_i$ and a half plane which has the boundary $l_i$ and does not contain $O$ lies in $S'$ (we get this using lemma and the aforementioned equivalence of the conditions of the problem). In such a way we rotate around the circle and get that all plane without the open disc centered at $O$ with radius $m$ lies in $S'$. (Indeed it is easy but nasty to make the above argument rigorous: one just have to choose $l'$ sufficiently close to the circle and $\epsilon$ sufficiently small. However intuitively it's clear what is going on). Hence an open disc centered at $O$ with radius $m$ lies in $S$. By the definition of $m$ no other points different from those lying on the boundary of the closed disc centered at $O$ with radius $m$ can lie in $S$. Let $T$ be any subset of this boundary and note that we can have $T\subset S$ and $S$ will satisfy the conditions of the problem. So $S$ is either all plane or a union $T\cup U$ where $U$ is any disc centered at $O$ without its boundary and $U$ is any subset of points of this boundary. Clearly all such $S$ satisfy the conditions of the problem.

hey this looks complicated! Any easier ways?

Firstly how does the lemma work? Surely you've only shown that the interior of the circle with diameter OA belongs to S? I think I have a nicer way of tackling the unbounded case. Suppose there exists a point P in the plane that is not in S. Then the line through P perpendicular to OP is also not in S (if it were, it would imply that P were in S, which it isn't). Also, for any point R of the plane on the other side of this line, one can choose a point Q on the line such that $\angle OQR = \pi/2$, so if R were in S, Q would be in S, again a contradiction, so the whole of the plane on the other side of this line would be excluded from S. We can also choose $Q\not =P$ on the line, draw the perpendicular to OQ through Q, and draw similar conclusions. If we do this a few times (so we next choose a point A such that AQ is perpendicular to OQ, etc.), we can form a closed polygon around O, which is a bound for S. Therefore, if S is unbounded, S must coincide with the plane. I didn't manage to find a solution for what S looks like when it's bounded though .

I beg your pardon, indeed the word radius in the statement of the lemma must be changed to diameter. But it's just a typo since afterwards I'm using the lemma properly.

I hope I haven't done any silly mistakes,because I spent quite some time writing down this proof(just when I was about to finish it my PC crashed and I had to start all over again :O ) Lemma 1: Suppose $A\in S$ then the closed disk of diameter $OA$ is contained in $S$. Proof: Take point $X$ in the disk, bring perpendicular to $OX$ at $X$ and let the section with the circle of diameter $OA$ be $Y$,then $X$ belongs to the circle of diameter $OY$(obviously $Y\in S$). Lemma 2: Suppose $A\in S$,then the open disk $(O,OA)$ is contained in $S$. First,we prove the following "sublemma": Given point $Z$ on circle $(O,OA)$ there is a point $X$ in the interior of the disk such that $\left | ZX \right |<\epsilon$ for every $\epsilon >0$ Proof: Split the angle $\measuredangle AOZ$ into $n$ equal parts by the lines $OA=\lambda_0,\lambda_{1}, \lambda_{2},...,\lambda_n=OZ$ and define the sequence of points $X_0=A$ and $X_{k+1}$ the projection of $X_{k}$ onto $OA_{k+1}$. All such $X_{k}$ belong to $S$ because $X_{k+1}$ belongs to the circle of diameter $OX_{k}$. Then I have $\left | OX_{k+1} \right |=\left | OX_{k} \right |\cos( \frac{\measuredangle AOZ}{n})$. Therefore $\left | OX_{n} \right |=\left | OA \right |(\cos( \frac{\measuredangle AOZ}{n}))^n$ and $\left | X_{n}Z \right |=\left | OA\right |(1-(\cos( \frac{\measuredangle AOZ}{n}))^n)$ which tends to $0$ as $n$ tends to infinity. Using the sublemma, I can have a point $X\in S$ at any radius $r<OA$ and by taking the union of the closed disks with diameter $OX$,where $X$ varies over all points at radius $r$, I get that the closed disk $(O,r)$ belongs to $S$ and since $OA$ is the supremum of such $r$ the open disk $(O,OA)$ belongs to $S$. Now, we take cases. If $S$ is unbounded, then it is the plane. Suppose there exists point $Z$ in the plane not contained in $S$, then there is a point with $OX>OZ$ and by lemma 2 $X\in S$,contradiction. If $S$ is bounded, then consider the set of points (not necessarily finite) $A$ that $OA=R$ is maximum distance(or the supremum of such distances). Then,by Lemma 2 $S$ contains the open disk $(0,R)$ and a set of points on the circle.