Here is my stackexcnage answer.

Another way of looking at this is to use the equivalence of categories between covering morphisms of a groupoid $P$ and actions of $P$ on sets. (Recall that a covering morphism $p:G \to P$ is a groupoid morphism having unique path lifting. Not necessarily unique path lifting gives a fibration of groupoids.) Given an operation of $P$ on a set $X$ then the corresponding covering morphism may be written $P \ltimes X$, an action groupoid, and thought of as a semidirect product because it is a special case of the semidirect product for an action of a groupoid $P$ on a groupoid $H$. For this one needs a morphism of groupoids $\omega: H \to Ob(P)$, where the latter is thought of as a discrete groupoid, and an element $w: x \to y$ in $P$ gives a morphism of groupoids $w_*: \omega^{-1}(x) \to \omega^{-1}(y)$. One has to be precise on conventions to get all this right, which I won't do here.

So a groupoid $G$ has a representation as an action groupoid whenever you are given a covering morphism $ G \to P$. This is closely related to Omar's answer, of course.

I'll add that more details of these ideas are in my book Topology and Groupoids.

Here is my stackexcnage answer.

Another way of looking at this is to use the equivalence of categories between covering morphisms of a groupoid $P$ and actions of $P$ on sets. (Recall that a covering morphism $p:G \to P$ is a groupoid morphism having unique path lifting. Not necessarily unique path lifting gives a fibration of groupoids.) Given an operation of $P$ on a set $X$ then the corresponding covering morphism may be written $P \ltimes X$, an action groupoid, and thought of as a semidirect product because it is a special case of the semidirect product for an action of a groupoid $P$ on a groupoid $H$. For this one needs a morphism of groupoids $\omega: H \to Ob(P)$, where the latter is thought of as a discrete groupoid, and an element $w: x \to y$ in $P$ gives a morphism of groupoids $w_*: \omega^{-1}(x) \to \omega^{-1}(y)$. One has to be precise on conventions to get all this right, which I won't do here.

So a groupoid $G$ has a representation as an action groupoid whenever you are given a covering morphism $ G \to P$. This is closely related to Omar's answer, of course.

I'll add that more details of these ideas are in my book Topology and Groupoids.

Addition: May 19, 2013 Here is a version of Sam's nice argument but in the language of covering morphisms.

If $G$ is a group then its universal cover $p: T \to G$ is a covering morphism of groupoids such that $T$ is connected and has trivial vertex groups; this is determined by the action of $G$ on itself by left multiplication. The set of objects of $T$ is bijective with the set $G$ and $T$ is the indiscrete groupoid (also called tree groupoid) on its set of objects. (Note that if $S$ is a generating set for $G$ then $p^{-1}(S)$ is a graph, namely the Cayley graph of $(G,S)$.)

Now let $A$ be a connected groupoid with $X$ as its set of objects. Let $T$ be the indiscrete groupoid on $X$. Then for any object $x$ of $A$, $A$ is isomorphic to $A(x) \times T$. But if $G$ is as above, then $$1 \times p: A(x) \times T \to A(x) \times G$$ is a covering morphism.

The next question is whether this argument can illuminate the case $A$ is a $\Gamma$-groupoid.