In Renault's A Groupoid Approach to $C^*$-Algebras, the following definition is given:
Let $G$ be a groupoid. A subset $S$ of $G$ will be called a $G$-set if the restrictions of $r$ [the range map] and $d$ [the domain map] to it are one-to-one. Equivalently, $S$ is a $G$-set iff $SS^{-1}$ and $S^{-1}S$ are subsets of $G^0$ [the unit space].
How does one see this equivalence? The backwards direction is clear to me: if $SS^{-1} \subseteq G^0$ then $r(st^{-1}) = st^{-1}$ for any $s,t \in S$ (when such multiplication is defined); but $r(st^{-1})=ss^{-1}$ so that if $r(s) = r(t)$ we have $st^{-1} = ss^{-1} = tt^{-1}$, which implies $s = t$. The case $S^{-1}S \subseteq G^0$ is treated similarly.
If $S$ is a $G$-set and $s,t\in S$ are such that $st^{-1}$ is defined, then $d(s)=r(t^{-1})=d(t)$. So by injectivity of $d$, $s=t$ so $st^{-1}\in G^0$. This shows $SS^{-1}\subseteq G^0,$ and $S^{-1}S\subseteq G^0$ is similar.