Let T = {(x,y) ∈ A x A | x ∈ Dom(R) and y ∈ Ran(R)}.

Clearly, if (x,y) ∈ R then (x,y) ∈ T. So R ⊆ T. Also, if (x,y),(y,z) ∈ T, then x ∈ Dom(R) since (x,y) ∈ T and similarly z ∈ Ran(R) since (y,z) ∈ T. Hence, (x,z) ∈ T, so T is transitive. Therefore, T ∈ M and so (x,y) ∈ T, which in turn implies x ∈ Dom(R).

Whew! I've proven the other subset direction and it is much easier. I'm sure the range equality is analogous and can use the same relation T. On the other hand, I'm not sure how to feel. I'm definitely happy the problem has been solved, but I feel I shouldn't have struggled as much with it (I usually post problems for help only after pondering them for a few hours with no avail).