For a finite graph $G$ it is quite easy to calculate the fundamental group. You just choose a maximal tree and then since trees are contractible, you get that the fundamental group of a finite graph is the free group $F_n$, where $n$ is the number of edges that is not contained in the maximal tree.
If the graph is directed, it is still possible to talk about loops (a cycle in a directed graph is a well-defined concept) and we can still multiply different loops by concatenation, so we can talk about a similar algebraic invariant. Since we do not have any inverses, the result is a monoid. Actually, since there will be no torsion and there are no homotopies, it will be a free monoid.
I wonder whether there is a easy way to calculate the fundamental semigroup (if the name is appropriate) of a directed graph. Clearly, the maximal tree trick does not work here.
We can find generators for the fundamental monoid as follows. Choose a vertex $v$ in the directed graph $G$ and construct two directed trees as follows. For the first tree, the the breadth-first search tree rooted at $v$. (This is our out-tree.) The second tree is chosen so that each vertex $w$ in the tree is joined by a directed path from $w$ to $v$; it is maximal with this property. This is our in-tree.
These two trees need not have the same vertex sets in general. We assume that that our directed graph is strongly connected, and under this assumption our trees are orientations of spanning trees of the underlying graph. Also each arc of the directed graph lies in a cycle.
Now any arc of $G$ that does not belong to one of the two trees determines a unique closed walk starting at $v$ - a walk in the out tree to the initial vertex of the arc, cross along the arc, then return to $v$ via the in-tree. It is immediate that the arcs not in either tree generate the fundamental monoid of $G$.