Show that $\mathbb{Z}$ is isomorphic to $n\mathbb{Z}$.
This is correct, with one odd comment: your "well-definedness" step is unnecessary.
Since this can be a subtle point, let me say a bit about what's going on here. "Well-definedness" shows up in the following situation:
Suppose we have a set $E$, and elements of $E$ correspond to equivalence classes under $\sim$ of some set $S$. Then we can define a function from $E$ by defining a function from $S$ instead ... provided that we separately show that $\sim$-equivalent elements of $S$ get sent to the same object.
I've deliberately written that informally. More precisely, but also maybe less intuitively:
Suppose $E=S/\sim$. Then if $f:S\rightarrow X$ has the property that $s_1\sim s_2\implies f(s_1)=f(s_2)$, then $f$ "induces" a function $\hat{f}: E\rightarrow X$ given by $\hat{f}([s])=f(s)$ for all $s\in S$.
Let's consider an example. We can think of a rational number as an equivalence class of elements of $\mathbb{Z}\times \mathbb{Z}_{\not=0}$, given by $$(a, b)\sim (c, d)\iff ad=bc.$$ For example, $(1, 2)\sim (2, 4)$ - that is, ${1\over 2}={2\over 4}$. Now consider the following:
Let $f:\mathbb{Q}\rightarrow\mathbb{Z}$ be defined by $$f({a\over b})=a+b.$$
(Here I suggestively write "${x\over y}$" instead of "$(x, y)$" for simplicity.) This is of course nonsense. What's really going on is:
I've defined a function $f:\mathbb{Z}\times\mathbb{Z}_{\not=0}\rightarrow\mathbb{Z}$ given by $f(a, b)=a+b$.
I've then implicitly claimed, without justification, that $f$ "respects $\sim$" - that is, $(a, b)\sim (c, d)\implies f(a, b)=f(c, d)$.
This second step of course is false: as observed above, $(1, 2)\sim (2, 4)$, but I leave it as an exercise to show that $1+2\not=2+4$.
Now what does this have to do with your situation? Well, nothing! That's because you've defined a function from $\mathbb{Z}$ directly; you're not defining a function on a set which $\mathbb{Z}$ is a quotient of, and then implicitly claiming that that function respects the equivalence relation. So there's no "well-definedness" claim here that needs to be proved.
(OK, there is actually a claim that needs to be proved here: you need to prove that the map $r\mapsto nr$ is actually a function from $\mathbb{Z}$ to $n\mathbb{Z}$, that is, each $x\in\mathbb{Z}$ gets sent to some $y\in n\mathbb{Z}$. This is essentially trivial - obviously $nr\in n\mathbb{Z}$, as long as $r\in\mathbb{Z}$ - but is necessary to prove in the sense of providing a completely rigorous proof. But that's not what you're talking about when you say "well-definedness," and also might not be something you need to explicitly write in this particular exercise.)