$\textbf{Lemma:}$ In $\triangle ABC$, $P$ is an interior point and $D=AP\cap BC$, $E=BP\cap AC$ and $F=CP\cap AB$. Let $Q\in AC$ such that $PQ\parallel BC$, $X=EF\cap AP$ and $Y=CX\cap BE$. Points $F,Y,Q$ are collinear if and only if $D$ is the midpoint of $BC$. $\textbf{Proof:}$ Just simple application of Ceva and Menelaus theorems. $\textbf{Notation:}$ Perpendicular bisector of $AB$ meet $AB$ at $M$ and $(ABC)$ at $D,E$ with $E$ in the same side of $\omega$ wrt $AB$. $\omega$ meet $AB, (ABC)$ at $T, R$ respectively and let $W$ be the center of $\omega$. $F=ET\cap (ABC)$, $K=ER\cap AB$ and the tangent at $(ABC)$ at $R$ meet $AB$ at $L$. $S$ is the antipode of $T$ in $\omega$. $P=S_1S_2\cap RD$. $Q=RF\cap AB$ Back to the problem, I will skip many details. $\textbf{Solution:}$ 1- Inversion at $D$ with radius $DA$ shows that $D,T,R$ are collinear and also $S_1, S_2$ remains fixed so they belongs to inversion circle. 2- Facts: * $E,S,R$ are collinear. * $D,F,K$ are collinear. * $L$ is the midpoint of $KT$ and $LF$ is tangent to $(ABC)$. 3- Inversion at $L$ with radius $LR$ interchanges $\omega_1$ and $\omega_2$, then $L,S_1,S_2$ are collinear and $T_1S_1S_2T_2$ is cyclic. 4- With the previous inversion $(FS_1S_2)$ is tangent to $(ABC)$, let this circle meet $AB$ at $S_1^*,S_2^*$ with $S_1^*$ closer to $B$. Now $RP$ is the polar of $L$ wrt $\omega$ then $FP$ is the polar of $L$ wrt $(FS_1S_2)$ then by Brocard $S_1^*S_1$, $S_2^*S_2$ and $FP$ concur. 5- $E,W,P,Q$ are collinear. From $E$, $(R,P,T,D)$ are harmonic and $TS\parallel DE$ then midpoint of $TS$ is on $EP$. In $KRTF$, $EP$ is the polar of $D$ then by Brocard $Q$ is on $EP$. 6- $S_1,T_1$ and $S_2,T_2$ meet at $S$. with the previous and using the lemma we have $(S_1^*,S_2^*) = (S_1, S_2)$ iff $S\in FP$ iff $O$ is the midpoint of $DE$ iff $AB$ is diameter of $(ABC)$. $\textbf{Remark:}$ Point $C$ is a red herring, everything can be deduce just from chord $AB$.

Point $C$ is irrelevent so we ignore it. We invert the entire problem around point $S$, the tangency point between $\omega_0$ and $\Gamma$, dropping $^*,'$, etc. We get the following new problem. Inverted Problem wrote: Let $\omega_0$ be a circle passing through $S,A,B$. Let $M$ be the midpoint of minor arc $\widehat{AB}$. Let $\ell$ denote the line tangent to $\omega_0$ at $M$. For $i=1,2$, let $\omega_i$ be the circle tangent to $\omega_0$ at $T_i$, $\ell$ at $S_i$, and $AB$. If $\omega_1$ and $\omega_2$ are distinct, prove that there exists a circle through $S_1,S_2,T_1,T_2$ and tangent to $\overline{AB}$ iff $\angle ASB = 90^{\circ}$. The advantage of this inversion is the obvious symmetry across the perpendicular bisector of $AB$; in particular, it already implies that $S_1,S_2,T_1,T_2$ are concyclic, and moreover, $\odot(S_1S_2T_1T_2)$, if tangent to $\Gamma$, must be tangent at the midpoint $K$ of $AB$. In addition, note by homothety that the lines $S_1T_1$ and $S_2T_2$ meet at the antipode $X$ of $M$. Now, we set $P=XS_1\cap AB$, and let $Q$ be the tangency point of $\omega_1$ and $AB$. We finish the problem by noting that \begin{align*} K\in\odot(S_1T_1S_2T_2) &\iff PT_1\cdot PS_1 = PK^2 \\ &\iff PQ = PK \\ &\iff P\text{ is midpoint of }\overline{S_1X}\\ &\iff \text{radius}(\omega_1) = \frac{\text{radius}(\omega_0)}{2} \\ &\iff K\text{ is the center of }\omega_0 \\ &\iff \angle ASB = 90^{\circ} \end{align*}