If $G$ is a group, $H$ is a subgroup of $G$, and $a,b\in G$ then the cosets $(ab)H=a(bH)$.
This obviously follows from associativity of group operation. But a formal proof should show first that $(ab)H\subseteq a(bH)$ and then $a(bH)\subseteq(ab)H$. However, I think an easier proof is this:
$$ (ab)H = \{(ab)h\mid h\in H\}$$ $$ = \{ a(bh)\mid h\in H\}$$ $$ = a \{bh \mid h\in H\}$$ $$ = a(bH).$$
Are we allowed to prove using set builder notation like this?
Your proof is fine. If you were to phrase it as a double inclusion proof, you would see that it is the exact same proof, just written slightly differently:
\begin{align} x \in (ab)H &\iff x = (ab)h && \text{for some }h \in H,\\ &\iff x = a(bh),\\ &\iff x \in a(bH). \end{align}