We first prove that $\triangle C M B \sim \triangle D B E$. Since $D$ and $E$ lie on opposite sides of $M B$, it follows that $\angle D B E=\angle D B M+\angle M B E=\angle D B M+\angle M D B=\angle C M B$ due to the given similarity and the exterior angle theorem. Secondly, it holds that

$$\frac{|D B|}{|B E|}=\frac{|M D|}{|M B|}=\frac{|C M|}{|M B|}$$

due to the given similarity and the fact that $M$ is the midpoint of $C D$. By SAS, it indeed follows that $\triangle C M B \sim \triangle D B E$. In particular, this implies that $\angle B E D = \angle M B C$. Therefore, $\angle A B D = \angle M B C$ if and only if $\angle A B D = \angle B E D$. By the tangent-secant theorem, this holds if and only if $A B$ is tangent to the circumcircle of $\triangle B D E$. The circle through $B$ and $D$ that is tangent to $A B$ is unique, with its center being the intersection of the perpendicular bisector of $B D$ and the line through $B$ perpendicular to $A B$. Thus, $A B$ is tangent to the circumcircle of $\triangle B D E$ if and only if $E$ lies on $\Omega$. $\square$