Solution It is well known that $S, C, M$ are collinear. Indeed, consider the dilation centered at $S$ that sends $\Gamma_1$ to $\Gamma_2$. Then the line $AB$ is sent to the line $l$ parallel to $AB$ and tangent to $\Gamma_2$, i.e. the line tangent to $\Gamma_2$ at $M$ (the midpoint of $\widehat{AB}$). Thus, this dilation sends $C$ (the points of tangency of the line $AB$ and $\Gamma_1$) to $M$ (the points of tangency of the line $l$ and $\Gamma_2$), from which it follows that $S, C, M$ are collinear.

By the power-of-point theorem, we have $AC \times CB = SC \times CM$. It suffices to show that
$$SC \times CM = 2r \times MN \quad \text{or} \quad \frac{SC}{2r} = \frac{MN}{CM}. \qquad \textcircled{1}$$
Set $\angle MCN = \alpha$. Then $\angle SCA = \alpha$. By the extended sine law, we have $\frac{SC}{2r} = \sin \alpha$. In the right triangle $MNC$, we also have $\sin \alpha = \frac{MN}{CM}$. Combining the last two equations, we obtain ①.

(We can also derive ① by observing that the triangles $MNC$ and $CDS$ are similar.)