In the comments of this question the directed graphs that can appear as commutative diagrams are axiomatized:
A graph is a combinatorial commutative diagram iff it's nonempty and such that any vertex is a part of a finite undirected cycle and such that any undirected cycle consists of two directed paths with the same start and end vertices.
Can it be proved for such graphs that:
Given two different edges directed from the same vertex, there are directed paths including these edges leading to a common vertex.
Not necessary, consider the directed graph on $A,B,C,B',C'$ with arcs $A\to B$, $B\to C$, $A\to C$, $A\to B'$, $B'\to C'$, and $A\to C'$. This directed graph is a counterexample to your claim since $A\to B$ and $A\to B'$ do not lead to a common vertex.
P.S. Combinatoric is not a word, combinatorial is.