Does this work? We divide the solution into two parts. One is the part where we examine the properties of $ABCD$ and the second one, where we present a construction. $\textbf{Observations:}$ Let $M$ be the midpoint of the diagonal $AC$ and $N$ be the midpoint of the diagonal $BD$. Let the incircle of $ABCD$ touch the sides $AB,BC,CD,DA$ at points $X,Y,Z,W$ respectively. Let $E$ and $F$ be the midpoints of $XZ$ and $YW$ respectively. Let $G=AB\cap CD$ and $H=AD\cap BC$. Denote by $V$ the intersection of $AC$ and $BD$. The following results hold: $XZ\perp YW$ Link 1 $V\in XZ, V\in YW$ Link 2 $V\in IO$ Link 3 $I\in MN$ Link 2 $EF\perp MN$ Link 2 $G\in EI$, $H\in FI$ (obvious) $V$ is the orthocenter of $\triangle GOH$ Link 2 The first result can be proved quickly: $$180^{\circ}-\angle YVZ=\angle VYZ+\angle VZY=\angle WYZ+\angle XZY=$$$$=\frac{180^{\circ}-\angle WDZ}{2}+\frac{180^{\circ}-\angle XBY}{2}=180^{\circ}-\frac{\angle ADC+\angle ABC}{2}=90^{\circ}$$$$\Longrightarrow \angle YVZ=90^{\circ}\Longrightarrow XZ\perp YW$$The sixth bullet point is obvious because $G$ is the intersection of the tangents at $X$ and $Z$ to $(XYZW)$, $E$ is the midpoint of $XZ$ and $I$ is the center of $(XYZW)$, so points $G,E,I$ lie on the perpendicular bisector of $XZ$, so $G\in ET$ and analogically $H\in FI$. The rest of the results would take a lot of time to compile and write out, so I've cited them and included links with their statements and proofs for shortness. $\textbf{Construction:}$ Let's construct the line $\ell$ through $M$, perpendicular to $OM$. This is line $AC$ as $OM\perp AC$, because $M$ is the midpoint of the chord $AC$ in $(ABCD)$, which has center $O$. Now we know that $V=OI\cap AC$, so if we intersect $\ell$ with $OI$ we'll construct point $V$. Now, we know that $ON\perp BD\Longrightarrow ON\perp NV$, but also from the third bullet point in $\textbf{Observations}$ we have that $I\in MN$, so if we construct the circle with diameter $OV$ and intersect $MI$ with this circle for the second time, then that would be point $N$, so we can construct point $N$. Now, we know that $XZ\perp YZ$ and $IE\perp XZ$ and $IF\perp YW$, so $IFVE$ is a rectangle, so the midpoint of $EF$ and $IV$ are the same point and $IEVF$ is cyclic. We also know that $EF\perp MN$ from the fifth bullet point, so if we construct $U$, the midpoint of $IV$, construct the circle $k(U,IU)$ and construct the line $g$ through $U$, perpendicular to $MN$, then the intersection points of $g$ and $k$ are points $E$ and $F$ from what we've mentioned above.Therefore we've constructed $E$ and $F$. Now we can construct points $G$ and $H$: we know that $V$ is the orthocenter of $\triangle GOH$ by the seventh bullet point, so $G,H$ are uniquely determined on $\overline{IE}$ and $\overline{IF}$. We will write this as a lemma, because it is not trivial. $\textbf{Lemma}$ We can construct points $G$ and $H$ with the points $O,I,V,E,F$. $\textit{Proof:}$ We will only need $E$ and $F$ to construct the lines $\overline{IE}$ and $\overline{IF}$, let's call them $t_{e}$ and $t_{f}$ respectively. We mentioned that $G\in IE=t_{e}$ and $H\in IF=t_{f}$. Let $R=OV\cap GH$ and $Q=HV\cap OG$. We mentioned that $V$ is the orthocenter of $\triangle GOH$, so $HQ\perp OG$ and $OR\perp GH$. Let $OI=x$, $IV=y$, $VR=z$, $\angle VIF=\alpha$ (we mentioned that $IFVE$ is a rectangle, so $\angle VIE=90^{\circ}-\alpha$). We have that: $$RH=\tan\angle RIH\cdot RI=(y+z)\tan\alpha$$$$GR=\tan\angle RIG\cdot RI=(y+z)\tan(90^{\circ}-\alpha)=\frac{(y+z)}{\tan\alpha}$$$$GH^2=(GR+RH)^2=(y+z)^2\left(\tan^2\alpha+2+\frac{1}{\tan^2\alpha}\right)$$$$OH^2=RH^2+OR^2=\tan^2\alpha(y+z)^2+(x+y+z)^2$$$$GV^2=GR^2+RV^2=\frac{(y+z)^2}{\tan^2\alpha}+z^2$$$$OV^2=(x+y)^2$$$$HV\perp OG\Longrightarrow GH^2+OV^2=GV^2+OH^2$$$$\Longrightarrow (y+z)^2\left(\tan^2\alpha+2+\frac{1}{\tan^2\alpha}\right)+(x+y)^2=\frac{(y+z)^2}{\tan^2\alpha}+z^2+\tan^2\alpha(y+z)^2+(x+y+z)^2$$$$\Longrightarrow 2(y+z)^2+(x+y)^2=z^2+(x+y+z)^2\Longrightarrow 2y^2+2yz=2zx\Longrightarrow \boxed{z=\frac{y^2}{x-y}}$$However, we have constructed points $O,I,V$, so we know the values of $x$ and $y$, we have $t_{e}$ and $t_{f}$ so we can construct $R$ by constructing a segment $VR'$ with length $\frac{y^2}{x-y}$ on line $\overline{OI}$, so that $V\in R'I$. From what we've shown $R'\equiv R$, so we have constructed $R$. Now, if we intersect the line through $R$, perpendicular to $OR$ with $t_{e}$ and $t_{f}$ we'll get $G$ and $H$ respectively, so we've proven the lemma. $\textbf{Note:}$ Notice that we also know $\alpha$, but we don't even need that because the expressions, including $\alpha$ cancel out. Having constructed $G$ and $H$, construct the circle $c(G,\sqrt{GE\cdot GI})$. The intersection of this circle with the line $VE\equiv XZ$ are the points $X$ and $Z$ (Notice that $XE\perp IG$ and $IX\perp XG\Longrightarrow \triangle GXE\sim\triangle GIX\Longrightarrow GI\cdot GE=GX^2$ and analogically $GI\cdot GE=GZ^2$; the intersection of the circle $c$ and $\overline{XZ}$ (which we can construct because we have constructed $E$ and $I$, so $\overline{XZ}$ is the line through $E$, perpendicular to $IE$) are two points, but we showed that $X$ and $Z$ lie on $c$ and obviously lie on $\overline{XZ}$, so the two intersection points are $X$ and $Z$). Analogically, if we construct the circle $t(H,\sqrt{HI\cdot HF})$ and intersect it with the line $VF\equiv YW$ we would get points $Y$ and $W$. Now we con construct $(XYZW)$ because we have constructed the points $X,Y,Z,W$ and have $I$. We can construct points $A,B,C,D$ as the intersection points of the tangents to $(XYZW)$ at $X,Y,Z,W$, so we have restored the quadrilateral. In more detail, if $l_{x}$, $l_{y}$, $l_{z}$, $l_{w}$ are the tangents to $(XYZW)$ at $X,Y,Z,W$ respectively, then: $$A=l_{x}\cap l_{w},\quad B=l_{x}\cap l_{y},\quad C=l_{y}\cap l_{z},\quad D=l_{z}\cap l_{w}$$One might say that in constructing $X,Y,Z,W$ we can't distinguish $X$ from $Z$ and $Y$ from $W$, but this is irrelevant, because if we were to change $l_{x}$ and $l_{z}$ (for example), the intersection points mentioned above won't change, that is, $A$ would be $l_{z}\cap l_{w}$ in this case for example, but we would have constructed the $4$ vertices of $ABCD$, so we've restored it.

Let $ABCD$ be the quadrilateral, and suppose $M$ is the midpoint of $BD$. Let $P$ be the intersection of the diagonals $AC,BD$. It is known that $IO$ passes through $P$. See here for a reference. We can therefore locate $P$ by intersecting $IO$ and the line through $M$ perpendicular to $OM$. Let $N$ be the midpoint of diagonal $AC$. By Newton's theorem for tangential quadrilaterals, $M,I,N$ are collinear. Then since $\angle ONP = 90^\circ$, we can locate $N$ by considering the intersection other than $M$ of line $MI$ and the circle with diameter $OP$. We will now use the fact that \[ \frac{AP}{CP} = \frac{AI^2}{CI^2}. \]This result is from Crux 2008 (see pg 50 of the link). [asy][asy] pair pole(pair A, pair B) { return 2*A*B/(A+B); } pair W = dir(42), X = dir(158), Y = dir(-128), Z = -W*Y/X, I = (0, 0), A = pole(W, X), B = pole(X, Y), C = pole(Y, Z), D = pole(Z, W), P = extension(A, C, B, D), M = (B+D)/2, N = (A+C)/2, O = circumcenter(A, B, D), Q = foot(I, A, C); draw(A--B--C--D--cycle, red); draw(circumcircle(A, B, D), heavymagenta); draw(A--C^^B--D, mediumred); draw(M--N^^O--P, royalblue); draw(unitcircle, heavycyan); draw(C--I--A^^I--Q, gray); string[] names = {"$A$", "$B$", "$C$", "$D$", "$M$", "$N$", "$I$", "$P$", "$Q$", "$O$"}; pair[] points = {A, B, C, D, M, N, I, P, Q, O}; pair[] ll = {A, B, C, D, M, dir(150), dir(0), dir(230), Q, O}; for (int i=0; i<names.length; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy] Let $Q$ be the foot of $I$ onto $AC$, and put $h=IQ$, $d=NQ$, and $c=AN=NC$. We seek to locate $A$ and $C$ along the line through $N$ perpendicular to $ON$. To this end, put $x=AN=CN$. By the length relation and the Pythagorean theorem, we have \begin{align*} \frac{x+c}{x-c} &= \frac{(x+d)^2+h^2}{(x-d)^2+h^2} \\ \implies (x+c)(x^2-2dx+d^2+h^2) &= (x-c)(x^2+2dx+d^2+h^2) \\ \implies (4d-2c)x^2 &= 2c(d^2+h^2) \\ \implies x = \sqrt{\frac{c(d^2+h^2)}{2d-c}} &= NI \sqrt{\frac{c}{2d-c}} = \sqrt{\frac{NI\cdot c}{2d-c} \cdot NI}. \end{align*}Using the last expression, it is clear how to construct $x$.