We first prove that $\triangle CMB \sim \triangle DBE$. Since $D$ and $E$ lie on opposite sides of $MB$, it holds that $\angle DBE = \angle DBM + \angle MBE = \angle DBM + \angle MDB = \angle CMB$ because of the given similarity and the exterior angle theorem. Moreover, it holds that
$$\frac{|DB|}{|BE|} = \frac{|MD|}{|MB|} = \frac{|CM|}{|MB|}$$
because of the similarity defining $E$ and the fact that $M$ is the midpoint of $CD$. It now follows that $\triangle CMB \sim \triangle DBE$ (sas). In particular, it follows that $\angle BED = \angle MBC$. Therefore $\angle ABD = \angle MBC$ if and only if $\angle ABD = \angle BED$. By the inscribed angle theorem (tangent case), this holds if and only if $AB$ is tangent to the circumcircle of $\triangle BDE$. The circle through $B$ and $D$ tangent to $AB$ is unique, and has as centre the intersection of the perpendicular bisector of $BD$ and the line through $B$ perpendicular to $AB$. So $AB$ is tangent to the circumcircle of $\triangle BDE$ if and only if $E$ lies on $\Omega$. $\square$