$(\Rightarrow )$ Suppose $CA=CB$. Take inversion with center $C$ radius $CA$. $\omega$ become circle passthrough $A,B$. $\omega_1$ become circle tangent to $AB,\omega'$ touch $\omega$ at $T'$. $\omega_2$ become circle tangent to $AB,\omega'$ touch $\omega$ at $S'$. Such that $T',S',C$ collinear and $\omega_1,\omega_2$ on the other side of $AB$ and both inside $\omega'$. By well-known lemma, we get $S'E' \cap \omega'$ at $M$, then $M$ is midarc $AB$ of $\omega'$. And $D'T' \cap \omega'$ at $N$, then $N$ is midarc $AB$ of $\omega'$, where $m\neq N$. Then we have $MN$ passthrough $C$. We want to proof $CE'=CD'$ (will lead to $CD=CE$ by $CD\cdot CD'=CE\cdot CE'$). Which is follow from Butterfly's Theorem.

Nice problem.

Let the tangent at $ S $ intersect $ AB $ at $ X $ and the tangent at $ T $ intersect $ AB $ at $ Y $. Note that $ XE^2=XA\cdot XB$ is equivalent with $ (XE-XA)AE=XA\cdot AB $ and similarly, $ YD^2=YB\cdot YA $ is equivalent with $ BD(YE-YB)=BY\cdot BA $, so $AE=BD $ iff $ \frac{XE}{XA}=\frac{YD}{YB} $. On the other hand, $ \frac{XE}{XA}=\frac{XS}{XA}=\frac{sin \angle SAC}{sin \angle CBS} $ and $ \frac{YD}{YB}=\frac{YT}{YB}=\frac{sin \angle ASC}{sin \angle BSC} $, so the two ratios are equal iff $ \frac{sin \angle SAC}{sin \angle ASC}=\frac{sin \angle CBS}{sin \angle BSC} $ or iff $ AC=BC $

For the necessary case, considering the homothety $\gamma_1\to\omega$, $F=DT\cap\omega$ is the midpoint of $\widehat{AB}$, and analogously, $G=ES\cap\omega$ is the midpoint of the other $\widehat{AB}$. Then we are done by the Butterfly theorem.

It was a problem in 2015 Italia MO. Both cases are trivial by using laws of sines.