Answer: no.
Since $(\angle ABC + \angle BCD) + (\angle BAD + \angle CDA) = 360^\circ$, one of these summands is not less than $180^\circ$. Without loss of generality, assume that $\angle ABC + \angle BCD \ge 180^\circ$. Denote the reflection of the triangle $MCD$ about $M$ by $MBD_1$. The inequality
$$\angle ABM + \angle MBE = \angle ABC + \angle BCD \ge 180^\circ$$
implies that the point $B$ lies either on the segment $AE$ or inside the triangle $AME$. In both cases $AB + BE < AM + ME$, which implies $AB + CD < AM + DM$.