Theorem: Suppose $x \in \mathbb{Q}$ with $x \neq 0$ and $y \in \mathbb{R - Q}$. Then $xy$ is irrational.
Proof. We proceed by contradiction. Suppose $xy \in \mathbb{Q}$. Then $xy = \frac{a}{b}$, where $a, b \in \mathbb{Z}, b \neq 0$. By hypothesis $x$ is rational, so we can write $x = \frac{c}{d}$, $c, d \in \mathbb{Z}$ and $d \neq 0$. Then $$y = \frac{a}{bx} = \frac{ad}{bc}$$
By closure of integer multiplication, $ad = m \in \mathbb{Z}$ and $bc = n \in \mathbb{Z}$. Then $y = \frac{m}{n}$, so $y \in \mathbb{Q}$.
But this is a contradiction, since by hypothesis $y \in \mathbb{R - Q}$.
Then our assumption that $xy$ is rational must be false. Then $xy$ is irrational. $\square$.
Discussion
The proof as written hinges on the fact that $xy = \frac{a}{b} \leftrightarrow y = \frac{a}{bx}$.
It feels a bit silly to ask, but is it reasonable to assume that (given "basic facts" about the rationals and integers such that one would have at the beginning of an introductory analysis course) I can "divide/multiply both sides" to obtain this result, and also the next result that $\frac{a}{b\frac{c}{d}} = \frac{ad}{bc}$?
Yes, this makes sense. When working within the rational numbers, or the real numbers, or the complex numbers, you are free to divide both sides of an equality with any non-zero number, including unknown / arbitrary non-zero numbers like $x$.
Technically, by the formal definition of fractions, your two equalities are both just different ways of writing $$ bxy=a $$ Since none of the fractions involved have zero in the denominator, you are in the clear.