Elementary approach to proving that a group of order 9 is Abelian

Let $G$ be a group of order $9$. If there's an element of order $9$, the group is cyclic, so assume all nonidentity elements have order $3$.

Let $x$ be a nonidentity element and let $y\in G-\langle x\rangle$. Then the elements of $G$ are $x^iy^j$, where $i, j\in \left\{0, 1, 2\right\}$. We are done if we show $xy=yx$. Consider $yx$. It must equal one of the $x^iy^j$. By cancellation, both $i$ and $j$ must be nonzero.

If $yx=x^2y$, then $yxy^{-1}=x^2$. This means that $y^3xy^{-3}=x^8=x^2\neq x$, a contradiction since $y^3=1$. Similarly, $yx\neq xy^2$. Now, if $yx=x^2y^2$, then $yx=x^{-1}y^{-1}=(yx)^{-1}$, which means $yx$ has order $1$ or $2$, clearly impossible.