For Hausdorff topological groups, the set $\{e\}$ containing only the identity is closed. This is because Hausdorff implies T1 which implies singletons are closed.
For topological groupoids, defined such that composition (of composable pairs) and inversion are continuous, the unit space $G^{0}=\{uu^{-1} : u\in G\}$ is also closed. The problem for me is how does one prove this? The text I've read quoted it without proof and I can't generalize the argument from the groups case.
Since you say that $G$ is a Hausdorff groupoid in the title, I believe that you assume that $G$ is Hausdorff.
The map $\delta \colon G \to G \times G, g \mapsto (g, gg^{-1})$ is continuous. The diagonal $\Delta = \{(g,g) \mid g \in G\} \subseteq G \times G$ is closed since $G$ is Hausdorff and $G^0 = \delta^{-1}(\Delta)$ is the pre-image of $\Delta$ with respect to $\delta$ since $g = gg^{-1}$ implies that $g$ is a unit and every unit satisfies $u = uu^{-1}$.