Let $O_1,O_2$ be the centers of $\omega_1,\omega_2.$ Assume that $X_1X_2$ passes through $M.$ If $U \equiv X_1Y_1 \cap O_1O_2,$ then $BS_1 \perp O_1O_2$ is the polar of $U$ WRT $\omega_1$ $\Longrightarrow$ Pencil $M(X_1,Y_1,S_1,U)$ is harmonic. Since $MS_1 \perp MU,$ then it follows that $AB,O_1O_2$ bisect $\angle X_1MY_1,$ i.e. $MY_1$ is the reflection of $X_1X_2$ across $AB.$ By similar reasoning, $MY_2$ is the reflection of $X_1X_2$ across $AB$ $\Longrightarrow$ $Y_1Y_2$ passes through $M.$

$ O_{2}MS_{2}X_{2} $ and $ MS_{1}X_{1}O_{1} $ are cyclic quadrilaterals because $ O_{2}M $ is perpendicular on $ AB $ and $ O_{2}X_{2} $ is perpendicular on $ S_{2}X_{2} $ and for the another is a similar proof. because $ X_{1},M,X_{2} $ are on a line then the angle $ S_{2}O_{2}X_{2} $ and the angle $ S_{1}O_{1}X_{1} $ are congruent. this also is the condition for the congruence of the angles $ S_{2}MY_{2} $ and $ S_{1}MY_{1} $ which means $ M $ in on the line $ Y_{1}Y_{2} $.

Solution: We consider the following assertion. The line $X_1X_2$ passes through $M$ if and only if the angle at which $\omega_1$ is visible from $S_1$ equals the angle by which $\omega_2$ is visible from $S_2$. Indeed, notice that $\angle O_2MS_2=90^{\circ}$ so the point $M$ lies on the circle passing through the points $S_2O_2X_2$ and we get that $\angle O_2MX_2=\angle O_2S_2X_2$. Similar arguments for $X_1$ yield that $X_1X_2$ are collinear if and only if $\angle O_2S_2X_2=\angle O_1S_1X_1$ which is the same as our assertion. Using the same argument for the other set of points, $Y_1$ and $Y_2$, we get the requested result.