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.