If $AB \parallel CD$ then the incircles of $AMCD$ and $BMDC$ have equal radii; now the problem conditions imply that the whole picture is symmetric about the perpendicular from $M$ to $O_1O_2$, and hence $ABCD$ is an isosceles trapezoid (or a rectangle). The conclusion in this case is true.

Now suppose that the lines $AB$ and $CD$ meet at a point $K$; we may assume that $A$ lies between $K$ and $B$. The points $O_1$ and $O_2$ lie on the bisector of the angle $BKC$. By the problem condition, this angle bisector forms equal angles with the lines $CM$ and $DM$; this yields $\angle DMK = \angle KCM$. Since $O_1$ and $O_2$ are the incenter of $KMC$ and an excenter of $KDM$, respectively, we have
$$\angle DO_2K = \frac{1}{2}\angle DMK = \frac{1}{2}\angle KCM = \angle DCO_1,$$
so the quadrilateral $CDO_1O_2$ is cyclic. Next, the same points are an excenter of $AKD$ and the incenter of $KBC$, respectively, so
$$\angle KAD = 2\angle KO_1D = 2\angle DCO_2 = \angle KCB;$$
this implies the desired cyclicity of the quadrilateral $ABCD$. $\square$