I've proved that ($\mathbb {Q}$ , +) is not isomorphic to ($\mathbb {Z}$ , +) as part of an assignment:
Let $f : \mathbb {Q} \rightarrow \mathbb {Z}$ and $a = f(1)$, then $a=f({1\over2}+{1\over2})=2f({1\over2})=3f({1\over3})=...$ and so on. Hence $\forall_{n\in\mathbb {Z}}$ $n$ divides $a$. Therefore, $a=0$ and so $f(n)=nf(1)=na=0$ $\forall_{n\in\mathbb {Z}}$ and so $f$ is clearly not a bijection.
And so I've been wondering if that same logic could apply to ($\mathbb {Q}$ , +) and any other group? Is there something that makes ($\mathbb {Z}$ , +) special here?
More like there's something special about $\mathbb{Q}$ here, namely that it is a divisible group while $\mathbb{Z}$ is not. You've pretty much said that yourself in the proof.
Being divisible is an algebraic invariant, meaning that if there is an isomorphism $G\to H$ then $G$ is divisible if and only if $H$ is.
Any other group? Clearly not, the property you've used is as I said that $\mathbb{Q}$ is divisible while $\mathbb{Z}$ is not. There are other examples of groups that are divisible and isomorphic to $\mathbb{Q}$, e.g. $r\mathbb{Q}=\{r\cdot q\ |\ q\in\mathbb{Q}\}$ for any irrational $r\in\mathbb{R}$.
On the other hand not all divisible groups are isomorphic, e.g. $\mathbb{Q}^2$ is not isomorphic to $\mathbb{Q}$ while both are divisible. A counterexample with numbers arises naturally from the above by embedding $\mathbb{Q}^2$ into $\mathbb{R}$, for example by considering $\mathbb{Q}+\pi\mathbb{Q}=\{q+\pi p\ |\ q,p\in\mathbb{Q}\}$ group or even $\mathbb{R}$ itself (although the last one is of higher cardinality).