While it's easy to show that something isn't commutative, it's not so easy to prove that something is anything – at least for me!
If we take a relatively simple example: $(\mathbb{Z}_{n}$ , $\times_{n})\space$ s.t. $\space a*b\space =$ $\space\overline{a\times_{n}b}$
It's definitely a binary operation: defined for any $(a,b) \in \mathbb{Z}_{n}$ , the image of any input $(a,b)$ is unique i.e. well-defined, and it's closed in $\mathbb{Z}_{n}$ – but how do you prove that is commutative?
My thinking: For $\space\overline{ab}\space$ to be equal to $\space\overline{ba}\space$ it must be that $ab \equiv ba \space(mod \space n) \iff ab-ba = nq$.
Therefore, if $\space\times_{n}\space$ is commutative then $\space ab - ba$ must equal $\overline{0}$. As $\overline{a}$ and $\overline{b}$ are residual classes in $\mathbb{Z}_{n}$ they are integers, then $ab - ba = 0$ and the result of zero is a factor of $n$, therefore $(\mathbb{Z}_{n}$ , $\times_{n})$ is commutative.
But, more generally, what steps should we take to prove commutativity of a binary operation?
$a \times_n b$ is the remainder of $ab$ divided by $n$, and likewise $b \times_n a$ is the remainder of $ba$ divided by $n$. Since $ab = ba$, they also have the same remainder when divided by $n$, so $a \times_n b = b \times_n a$.
In general, you have to show that $a \star b = b \star a$ without making any assumption about $a$ and $b$ beyond the fact that they are elements of the set under consideration, so that your argument will apply equally to all the elements of the set.