Let $A, B \in U(n)$ for $n \in \mathbb N^+$. Then we need to show that (ordinary) multiplication of $A, B$ satisfies the following,
$$(AB, n) = 1$$
which can be done using the Bézout's lemma for $A, B$ respectively, then multiplying and manipulating the result.
$$Ax + ny = 1,$$ $$Bx' + ny' = 1,$$
For some $x, x', y, y' \in \mathbb Z$.
$$(Ax + ny)(Bx'+ny') = 1$$ $$\Longrightarrow$$ $$[AB]u +[n]v = 1$$
which shows us that $AB$ and $n$ are relatively prime. For sake of readability, i didn't write the cumbersome calculations but instead of it used $u$ and $v$ as placeholders.
If we apply the division theorem to $AB$ with $n$ we will obtain the remainder $R$ which is same as the modular multiplication of $A$ and $B$. Respectively by the division theorem and Euclid's method we know that,
$$0 < R < n,$$ $$(AB, n) = (n, R) = (R, n) = 1.$$
So it has to be true that $R \in U(n)$, hence this finite group is closed under modular multiplication.
$\backsim$ QED $\backsim$