I'm quite new to modular arithmetic. Therefore, this question may be very silly. Sorry for this.
Suppose that I want to prove that: $$a \cdot b \equiv 0 \mod{c}.$$
If I know that
$$c \equiv 0 \mod{a},$$
then can I conclude that:
$$b \equiv 0 \mod{d} \Rightarrow a \cdot b \equiv 0 \mod{c},$$
where $c = a \cdot d$?
Yes, it's simply $\ \bbox[5px,border:1px solid #c00]{ad\mid ab\iff d\mid b}\,\ $ by cancelling or scaling by $\,a\neq 0,\ $ i.e.
$\ \, \dfrac{ab}{ad} = \dfrac{b}d,\ $ so $\,\ ad\mid ab\iff \dfrac{ab}{ad}\in \Bbb Z\iff \dfrac{b}{d}\in\Bbb Z\iff d\mid b$
Alternatively, eliminating fractions and emphasizing an equational view
$$ ad\mid ab\iff \exists x\!:\, ad\,x = ab \iff \exists x\!:\, d\,x = b\iff d\mid b $$
i.e. equationally: $\ \,ad\,x = ab\,$ has an integer root $x\!\iff\! d\,x = b\, $ has an integer root $x$
with very easy proof: cancelling $\,a\neq 0\,$ yields $(\Rightarrow)$, and scaling by $\,a\,$ yields $(\Leftarrow)$.
i.e. the operation of "scaling by $a$" is invertible if $a$ is cancellable, so scaling by $a$ and cancelling $a$ yield equivalent equations (same set of roots), so these operations preserve solvability.
This preservation of equations is well known in the special case that $a$ is invertible (e.g. $\,a\neq 0\,$ in a field like $\,\Bbb Q,\Bbb R,\Bbb C)$ when we normalize polynomials to be monic (lead coef $=1$) by scaling by $a^{-1}$.
Alternatively it can be viewed as a special case of congruence scaling / division / cancellation when written in the language of congruences (vs. divisibility).