An inversion with center $O$ and constant $r^2$ ($r$ is the radius of the circumcircle of quadrilateral), takes the circle with diameter $OS$ to the line $EF$ and the circumcircle of the triangle $OAM$ to the line $AM$. Furthermore, since $XO$ is the bisector of $\angle AXB$ and $OA = OB$, we get the quadrilateral $AXBO$ is cyclic and so $OE' \cdot OX = r^2$. Similarly, $OF' \cdot OY = r^2$. Therefore, under this inversion the circumcircle of triangle $OXY$ maps to the line $E'F'$. So it is enough to prove that the line $EF$, $E'F'$ and $AM$ are concurrent.

Denote by $\omega$, $\omega_E$ and $\omega_F$ the incircles of quadrilateral $ABCD$, triangle $EAB$ and triangle $FAD$, respectively. Now $E$ and $E'$ are the external and internal homothetic centers of $\omega_E$ and $\omega$, respectively. In the same manner $F$ and $F'$ are the external and internal homothetic centers of $\omega_F$ and $\omega$, respectively. Therefore, $EF$ and $E'F'$ meet at the internal homothetic center of $\omega_F$ and $\omega_E$, say $R$. On the other hand, $AM$ is the bisector of $\angle EAB = \angle FAD$ and so is the line of centers of $\omega_F$ and $\omega_E$. Thus, $R$ the internal homothetic center of $\omega_E$ and $\omega_F$ lies on $AM$. So we have proved that the lines $EF$, $E'F'$ and $AM$ meet at $K$, which completes the proof.