Claim restatement: Let $a_1, a_2, b_1, b_2 \in \mathbb{Z}$ and $n \in \mathbb{N}$, as well as $a_1 \equiv b_1 \mod{n}$ and $a_2 \equiv b_2 \mod{n}$. Then, $a_1a_2\equiv b_1 b_2 \mod{n}$
I always see the same proof for this.
Proof 1:
By the congruence relations, we get $$n | a_i-b_i $$ and we want to show that $$n|a_1a_2 - b_1 b_2$$
$$a_1a_2 - b_1b_2 = a_1a_2 - b_1a_2 + b_1a_2 -b_1b_2=a_2(a_1-b_1)+b_1(a_2-b_2)$$
and because $\mathbb{Z}_n$ is closed under addition, the result follows.
But there's another more straightforward proof, similar to that of closure under addition, that I never see anywhere.
Proof 2:
From the congruence relations, $\exists k_1, k_2 \in \mathbb{Z}$ such that $$a_i = k_i n+b_i$$
Multiplying, we get
$$a_1a_2=(k_1n+b_1)(k_2n+b_2)=k_1k_2n^2+b_1k_2n+b_2k_1n+b_1b_2$$ $$a_1a_2-b_1b_2=k_1k_2n^2+b_1k_2n+b_2k_1n=n(k_1k_2n+b_1k_2+b_2k_1)$$
$$\frac{a_1a_2-b_1b_2}{n} = k_1k_2n+b_1k_2+b_2k_1 \in \mathbb{Z}$$ By closure in $\mathbb{Z}$ under addition and multiplication.
Thus, clearly $$n|a_1a_2-b_1b_2$$
But I never see this second proof. What makes the first one more desirable?
For one thing, the first does not use division anywhere, which is good because apriori, modular arithmetic makes sense outside of $\mathbb Q$, i.e: without reference to division at all.
Another reason, is that yours is a restatement in disguise, without reference to extraneous variables.
On the other hand, your proof is over at the penultimate line, since $n \mid (a_1a_2-b_1b_2)$. I'm not sure what being in $\mathbb Z$ has to do with this, so I don't really see the necessity of having those lines, rather than just factoring $n$ out immediately.