I want to prove that $(\mathbb{Z}/n\mathbb{Z},+)$ is a group. I have proved it myself but, as an unexperienced student, I'd like to check with you if my reasoning is correct. I have already proved that the operations: $$ [a]+[b]:= [a+b] \quad \mbox{and} \quad [a][b]:=[ab]$$ are well-defined, and this proof is correct since I followed my algebra book. Now, to prove $(\mathbb{Z}/n\mathbb{Z},+)$ is a group, this is what I've done:
(1) $([a]+[b])+[c] = [a+b]+[c] = [a+b+c] = [a]+[b+c] = [a]+([b]+[c])$, proving associativity of the sum;
(2) The identity element on this group is $[0]$ since $[a]+[0] = [a+0] = [a] = [0+a] = [0]+[a]$;
(3) If $a \in \mathbb{Z}$, then $[-a]$ is the inverse element of $[a]$ in $\mathbb{Z}/n\mathbb{Z}$ so that $[a]+[-a] = [a+(-a)] = [0]$.
I think this reasoning is okay, but I wonder if when taking $a \in \mathbb{Z}$ and its equivalence class $[a]\in \mathbb{Z}/n\mathbb{Z}$ I must first state that this $[a]$ is equivalent to one of these sets $[0],[1],...,[n-1]$ and then prove the result for this element instead of $[a]$ itself, but I think I don't need to do this because I proved the sum is well-defined so it does not depend on the choices of representatives. Is my reasoning correct?
Your proof is correct.
However, for (1), I would write this:
$$\begin{align} ([a]+[b])+[c] &= [a+b]+[c] \\ &=\color{blue}{[(a+b)+c]}\\ &=\color{blue}{[a+(b+c)]}\\ &= [a]+[b+c] \\ &= [a]+([b]+[c]). \end{align}$$
It is not necessary to state that $[a]\in\{[0],[1],\dots,[n-1]\}$ for each, or any, $a\in\Bbb Z$.
Also, you did not use $[a][b]:=[ab]$. Are you sure you mean "group" and not "ring"?