QUESTION: Let $S$ be a subset of $G$ such that the identity element $1 \in S$. Assume that the subsets $aS := \{as \mid s \in S\} \subseteq G$ for $a \in G$ form a partition of $G$. Prove that $S$ is a subgroup of $G$.
This is my solution thus far. Can someone check if it is remotely correct and where I can improve?
Your equation involving $\{1\}$ is unnecessary and seems wrong. (Did you mean $\{x\}\cup\dots$ instead of $\{1\}\cup\dots$ ?) For that step, I'd just write "we have $x=x1\in xS$". Also, you might want to note that $S=1S$ is of the form $aS$ in order to apply the fact about the partition. Besides that, LGTM.