The definition of a Lie groupoid action given in Sébastien Racanière's notes (up to notation) says that the action of a Lie groupoid $\mathcal{G} \rightrightarrows M$ on a smooth manifold $Q$ consists of two things:
- a smooth map $J:Q \to M$;
- a smooth map $\mathcal{G}\, {_{\sf s}}\!\times_J \!Q \to Q$ such that $J(g \cdot q) = {\sf t}(g)$ and $g\cdot (h\cdot q) = (gh)\cdot q$;
This makes perfect sense, since in the same way one cannot multiply two arbitrary elements of $\mathcal{G}$ (this is controlled by source ${\sf s}$ and target ${\sf t}$), an element of $\mathcal{G}$ would not be able to act on the entire manifold $Q$ (this is controlled by the map $J$). Moreover, axiom 1 together with the assumptions ${\sf t}(h) = {\sf s}(g)$ and $J(q) = {\rm s}(h)$ indeed ensures that both sides of $g\cdot(h\cdot q) = (gh)\cdot q$ are defined.
Now, I would expect that if $x \in M$ is a unit, then $x\cdot q = q$. However, since $x \in \mathcal{G}$ is a unit if and only if $x^2=x$, the best I can conclude is that $x$ fixes $x \cdot q$, for all $q \in Q$ such that $J(q) = x$.
Is this the best analogy we can get with the group axiom $e \cdot q = q$, for an action $G \circlearrowright Q$?