If $S$ is the intersection of the common external tangents, then $S$ is the exsimilicenter of $(AEB), (AED)$. Let $O_B, O_D$ be their respective centers; then by spiral similarity, $\triangle{AO_BO_D} \sim \triangle{ABD}$, so $(AES)$ is the Circle of Apollonius containing all points with distances to $O_B, O_D$ in the ratio $AB : AD$. Thus, it suffices to show that $\frac{CO_D}{CO_B} = \frac{CD}{CB}$. We claim that $\triangle{CO_DB} \sim \triangle{CO_BD}$; this would suffice. Note that if $M$ is the midpoint of $AE$ then $M$ lies on the perpendicular bisector of $BD$, so $BM = DM$, and $\angle{CBO_D} = 90^{\circ} - \angle{DBM}$, so $\angle{CBO_D} = \angle{CDO_B}$. Thus, it suffices to show that $\frac{CB}{BO_D} = \frac{CD}{CO_B}$, which is equivalent to $\frac{BO_D}{DO_B} = \frac{BC}{CD}$. Since this is also the ratio of the radii of the two circles, by Power of a Point it suffices to show that $$\frac{Pow_{O_D}(B)}{Pow_{(O_B)}(D)} = \left(\frac{BC}{CD}\right)^2,$$which is clearly true since this is equal to $\frac{BE \cdot BD}{DE \cdot BD}$, and $\frac{BE}{DE} = \left(\frac{BC}{CD}\right)^2$ by similar triangles. $\square$