How to show that $(n\mathbb Z+a)(n\mathbb Z+b) = n\mathbb Z + (ab)$?

Question

I am studying the equivalence classes of the relation congruence $\operatorname{mod}n$ on $\mathbb Z$. I define for any $A, B\subseteq\mathbb Z$, \begin{align*} A+B & := \{a+b : a\in A, b\in B\}\text{, and}\tag{1}\\ AB & := \{ab : a\in A, b\in B\}\text{.}\tag{2} \end{align*}

With these definitions in hand, I can show that$^1$ $$ (n\mathbb Z + a) + (n\mathbb Z+b) = n\mathbb Z+(a+b)\text{.}\tag{a} $$

But I am struggling to show that $$ (n\mathbb Z+a)(n\mathbb Z+b) = n\mathbb Z + (ab)\text{.}\tag{b} $$

I have figured out that I will be done if I can prove that for an $n, a, b, k\in\mathbb Z$, there exist $k_1, k_2$ such that $k = nk_1 k_2 + k_1 a + k_2 b$. But I am stuck here. Induction doesn't seem to help.

Questions:

1. Can someone help with the proof of $(\text{b})$? (I can show that the LHS is a subset of the RHS, not the other way around.) Or give a counterexample?
2. If it is false in general, is there any other definition of $AB$ so that the result follows? Or should I really just take $(\text{a})$ and $(\text{b})$ as my (less general) definitions?

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$^1$ $n\mathbb Z+a :=\{nk+a : k\in\mathbb Z\}$. It is easily shown that $n\mathbb Z+a$ is the equivalence class (also called congruence class) containing $a$.

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Thanks to Conifold for his comment. According to definition $(2)$, I'll have $(2\mathbb Z)(2\mathbb Z) = 4\mathbb Z$, whereas we want the RHS to be $2\mathbb Z$ according to $(\text{b})$. Hence, $(2)$ and $(\text{b})$ are not compatible.

Further, see his comments on how to define $AB$ when $A$, $B$ are equivalence classes.