Let $U \equiv AX \cap BX',$ $V \equiv AY \cap BY'$ and let $AX',AY'$ cut $\Gamma$ again at $N,M,$ respectively. Clearly, $A$ is center of inversion taking $\Gamma$ into $l$ and this carries $\odot(AXY)$ and $\odot(AX'Y')$ into $UV$ and $MN,$ respectively. Thus, it's enough to show that $l,MN,UV$ concur. From the inversion with center $A,$ it follows that $M,N,X',Y'$ are concyclic and since $B$ is also center of inversion taking $\Gamma$ into $l,$ then $U,V,X',Y'$ are concyclic. Thus $UV,MN,l$ are pairwise radical axes of $\Gamma,$ $\odot(MNX'Y'),$ $\odot(UVX'Y')$ concurring at their radical center, as desired.

The solution I wrote during the test was along the lines of this, but in more detail: Let $AX\cap BX'=X_0$, $AY\cap BY'=Y_0$, and $AB\cap l=I$ Notice that $YY_0BI$ and $XX_0BI$ are cyclic, then by PoP/Radical Lemma, $XX_0YY_0$ are cyclic. Similarly, we can prove that $X'Y'X_0Y_0$ is cyclic. Consider the radical center of $\Gamma$, $XYX_0Y_0$, and $AXY$. The radical center will be the intersection of $XY$ and $X_0Y_0$, Let this point be $C$. Then the radical axis of $\Gamma$ and $AXY$ is $AC$. Similarly, consider the radical center of $\Gamma$, $X'Y'X_0Y_0$, and $AX'Y'$, the radical center is the intersection of $X'Y'$ and $X_0Y_0$, which is again $C$. So the radical axis of $\Gamma$ and $AX'Y'$ is also $AC$ $\Gamma$, $AXY$ and $AX'Y'$ all share the same radical axis. Therefore they must all intersect on another point on $\Gamma$ other than $A$. The 3 circles are tangent when $X_0Y_0$ and $l$ don't intersect. Which is achieved when $X,Y$ are reflections across $AB$.