This is one of my first proofs about groups. Please feed back and criticise in every way (including style & language). Axiom names (see Wikipedia) are italicised. $e$ denotes the identity element.
Let $(G, \cdot)$ be a group. We assume that every element is its inverse. It remains to prove that our group is commutative. Non-trivially, $\textit{associativity}$ implies that parentheses are unnecessary. Therefore, we do not use parentheses, we will not use $\textit{associativity}$ explicitly.
By $\textit{identity element}$, $G \ne \emptyset$. Now, let $a, b \in G$. By assumption, $$aa = e \text{ and } bb = e. \quad \text{(I)}$$ By $\textit{closure}$, $ab \in G$. So, by assumption, $$abab = e.\quad \text{(II)}$$ It remains to prove that $ab = ba$. \begin{equation*} \begin{split} ab &= aeb && \quad\text{by }\textit{identity element} \\ &= aababb && \quad\text{by (II)} \\ &= ebabb && \quad\text{by (I)} \\ &= ebae && \quad\text{by (I)} \\ &= bae && \quad\text{by }\textit{identity element} \\ &= ba && \quad\text{by }\textit{identity element} \end{split} \end{equation*} QED
It is correct, however, you might see the way similarly as: $$a(ab)b=a^2b^2=e=(ab)^2=a(ba)b.$$