This question is more concerned with understanding the reasoning behind mathematical proving rather than explaining this specific proof. I understand why this proof works.
This is the proof of Euclid Lemma from Wikipedia:
This states that if $x$ and $y$ are relatively prime integers (i.e. they share no common divisors other than 1) there exist integers $r$ and $s$ such that
$rx+sy=1$. Let $a$ and $n$ be relatively prime, and assume that $n|ab$. By Bézout's identity, there are $r$ and $s$ making
$rn+sa=1$. Multiply both sides by $b$:
$rnb+sab=b$. The first term on the left is divisible by $n$, and the second term is divisible by $ab$, which by hypothesis is divisible by $n$. Therefore their sum, $b$, is also divisible by $n$.
I am mostly concerned with this statement:
Multiply both sides by $b$:
My question is why this multiplication by $b$ happens. Is it because the person proving this observed that multiplying the equation by $b$ he would get a sum of two multiples of $n$? Or is there a mathematical rule by which we know that the next step is multiplying by $b$?
I am aware that this question might seem trivial to people here, but I just want to make sure.
Yes, the "rule" is: $ $ an $\color{#c00}{\rm invertible}$ element is cancellable (simply by scaling by its inverse), and integers coprime to $\,n\,$ are invertible$\bmod n\,$ (by Bezout), see here. $ $ Therefore we have:
$\qquad\qquad\ \ \ \ n\mid ax\,\Rightarrow\, n\mid x\ \ $ if $\,\gcd(a,n) = 1,\, $ interpreted mod $n$ becomes
$\quad \iff\ \ \ ax\equiv 0\,\Rightarrow\, x\equiv 0\ $ if $\,\gcd(a,n) = 1.\ $ The coefficient $\,a\,$ is invertible by Bezout:
$\qquad\ \ \qquad \color{#c00}{sa\equiv 1}\,\,\ $ by $\ sa = 1 - rn\ $ for some $\,s,r,\,$ by Bezout identity for $\,\gcd(a,n) = 1$
$\qquad\, \Rightarrow\,\ \color{#c00}{sa}x \equiv s0\ $ by scaling 2nd last equation by $\,\color{#c00}s,\,$ valid by the Congruence Product Rule
$\qquad\, \Rightarrow\,\ \quad x\equiv 0\ $ by $\ \color{#c00}{sa\equiv 1}$
i.e. scale by $\,\color{#c00}{s\equiv a^{-1}}\,$ to cancel $\,\color{#c00}a\,$ from $\,\color{#c00}ax\equiv \color{#c00}a0\,$ to get $\,x\equiv 0\,$ (that's how to view it explicitly as cancelling $\,\color{#c00}a\,$ on both sides).
Remark $\ $ Thus reformulating the divisibility relation as arithmetical operations mod $\,n\,$ clarifies the arithmetical essence of the matter. In the same way many divisibility properties are greatly clarified and simplified when reformulated arithmetically in congruence form.