If we can prove that $S$ is $AA \cap BC$ then we will be finished, since $(S)$ and $\Gamma$ will be orthogonal and $\angle OMB= 90$. It is clear that $S$ is on $BC$ so let's try to prove that $SA$ is tangent to $\Gamma$. It is known that $AE$ is symmedian in $ABC$ thus $(A, E; B, C)=-1$ and because of the fact that $SA=SE$, we have $AE$ is the polar line of $S$ and we are done.

Let $E'$ be a point on $(ABC)$ such that $AE'$ is the $A$ symmedian in $\triangle ABC$. Note that lines $SM, BC$ are the perpendicular bisector of $AD$. $\measuredangle BME' = \measuredangle ACE' = \measuredangle AMB = \measuredangle BMD \implies E' \equiv E$. Also $\measuredangle AOE = 2 \measuredangle ACE = \measuredangle ACM + \measuredangle MCE = \measuredangle AME$ and because of the harmonic config $SA , SE$ are tangent to $(ABC) \implies \measuredangle ASE = 180 - 2\measuredangle EAC = 2 \measuredangle AEC = \measuredangle AOE$. $~ \blacksquare$

Let $H$ be the orthocenter of $ABC$. Because the reflection of $E$ over $BC$ lies on $MA$ and $(BHC)$, we know it coincides with the $A$-Humpty point, i.e. $X_a$. But it's well-known that the reflection of $X_a$ over $BC$ lies on the $A$-Symmedian, so $ABEC$ is harmonic. Now, define $S_1 = AA \cap BC \cap EE$. Thales' implies $AOMES_1$ is cyclic, so it suffices to show that $S_1$ is the circumcenter of $ADE$. Indeed, we have $S_1A = S_1D$ from symmetry and $S_1A = S_1E$ from common tangents, so the desired result follows readily. $\blacksquare$ Remark: We can reverse engineer the precise characterization of $S$ by realizing $BC$ is the perpendicular bisector of $AD$ and considering the circumcircle of $AOM$.