If $I$ is the incenter of $ABCD,$ then $E' \equiv IE \cap AD$ and $F' \equiv IF \cap AB.$ Condition $\tfrac{XA}{XD}=\tfrac{EA}{ED}$ means that $X$ is on the E-Apollonius circle of $\triangle EAD,$ which is orthogonal to any circle through $A,D,$ thus particularly orthogonal to the circumcircle $(O,R)$ of $ABCD$ $\Longrightarrow$ $R^2=OE' \cdot OX$ and similarly $R^2=OF' \cdot OY.$ Inversion WRT $(O,R)$ takes $X,Y$ into $E',F',$ takes the circle with diameter $\overline{OS}$ into $EF$ (as $EF$ is the polar of $S$ WRT $(O)$) and fixes $A,M.$ If $E'F',IA$ cut $EF$ at $T,K,$ then from the complete quadrilateral $EE'F'F,$ it follows that $(E,F,K,T)=-1.$ But since $AM,AI$ bisect $\angle BAD,$ we have $A(E,F,K,M)=-1$ $\Longrightarrow$ $M \in AT$ $\Longrightarrow$ $E'F',AM$ and $EF$ concur at $T$ $\Longrightarrow$ their inverses $\odot(OXY),$ $\odot(OAM)$ and the circle with diameter $\overline{OS}$ intersect at a second point, i.e. they are coaxal.

Nothing too different from above. TheOverlord wrote: $ABCD$ is a circumscribed and inscribed quadrilateral. $O$ is the circumcenter of the quadrilateral. $E,F$ and $S$ are the intersections of $AB,CD$ , $AD,BC$ and $AC,BD$ respectively. $E'$ and $F'$ are points on $AD$ and $AB$ such that $A\hat{E}E'=E'\hat{E}D$ and $A\hat{F}F'=F'\hat{F}B$. $X$ and $Y$ are points on $OE'$ and $OF'$ such that $\frac{XA}{XD}=\frac{EA}{ED}$ and $\frac{YA}{YB}=\frac{FA}{FB}$. $M$ is the midpoint of arc $BD$ of $(O)$ which contains $A$. Prove that the circumcircles of triangles $OXY$ and $OAM$ are coaxal with the circle with diameter $OS$. Let $\{I,O\}$ be the Incenter and Circumcenter of $ABCD$ respectively as $\frac{FA}{FB}=\frac{AF'}{F'B}$. So, by Angle -Bisector Theorem we get that $\overline{F-F'-I}$ and similarly $\overline{E-E'-I}$. Now as $\frac{YA}{YB}=\frac{FA}{FB}$. Hence, $Y\in F-\text{ Appolonius Circle } (\omega_1)$ of $\triangle AFB$. Let $\omega_1\cap AB=X_1$. So, $X_1\in\text{ Polar of }F'$ WRT $\odot(ABCD)$.So, By Self-Orthogonality Lemma $\omega_1\perp\odot(ABCD)$. So, $OY\cdot OF'=R^2$ where $R$ is the circumradius of $\odot(ABCD)$. Similarly $OX\cdot OE'=R^2$. Now Consider an Inversion $(\Psi)$ around $\odot(ABCD)$. $\Psi$ swaps $\{S,(\overline{O-I-S})\cap EF\}$ (by Brocard), $\{F',Y\},\{E',X\}$ and $\{A,M\}$ remains fixed. So, $\Psi$ swaps $\{\odot(OS),EF\},\{\odot(OAM),AM\},\{\odot(OXY),E'F'\}$. So, the Problem becomes equivalent as follows. Inverted Problem wrote: $ABCD$ be a bicentric quadrilateral with $I$ as the center of Inscribed Circle and $AB\cap CD=E$ and $AD\cap BC=F$. Let $\{IF\cap AB=F'\}$ and $\{IE\cap AD=E'\}$ and let $M$ be the midpoint of Minor Arc $BD$. Then $\{EF,AM,E'F'\}$ concurs. Let $EF\cap E'F'=T$ and $TA\cap\{\odot(ABCD),IE\}=\{M',K\}$ respectively. So, $-1=(E,E';K,I)\overset{A}{=}(B,D;M',AI\cap\odot(ABCD))$. Now notice that $AI\cap\odot(ABCD)$ is the Midpoint of $\widehat{BCD}$. Hence, $M'$ must be the midpoint of Minor Arc $BD$. Hence, $M'\equiv M$. Hence $\{EF,AM,E'F'\}$ are concurrent. So, Inverting back we get that $\{\odot(OXY),\odot(OAM),\odot(OS)\}$ are coaxial. $\blacksquare$