In Olympiad math context (mainly AoPS), I often see substitution $(x,\ y,\ z,\ w)=(ab,\ cd,\ ad,\ bc)$ under $xy=zw$. The following is a construction of $(a,b,c,d)$ which I came up with.
For a prime $p$ and nonzero integer $n$, we let $\nu_p(n)$ denote the largest integer $e$ with $p^e$ dividing $n$. For an arbitrary prime $p$, let $$(s,\ t,\ u,\ v)=(\nu_p(x),\ \nu_p(y),\ \nu_p(z),\ \nu_p(w))$$ We have $s+t=u+v$. Without loss of generality we may assume $s=\min\{s,t,u,v\}$. We let $(a,b,c,d)$ satisfy \begin{eqnarray} \nu_p(a)&=&s \\ \nu_p(b)&=&0 \\ \nu_p(c)&=&u \\ \nu_p(d)&=&v-s \end{eqnarray} Now we have $\nu_p(a)+\nu_p(b)=\nu_p(x)$, $\nu_p(c)+\nu_p(d)=\nu_p(y)$, etc. We repeat this operation and the construction is done.
The question is:
Does this substitution have a particular name that one can easily refer to?
I don’t know of a name for the substitution… but I do believe there’s an easier answer/proof than what you’re suggesting. (Coincidentally, I was just writing it up for a paper I’m preparing for publication soon! I’m using the restricted version where $x,y,z,w$ are positive, but the full version is proven in a similar manner.)
Proposition. If $a$, $b$, $c$, and $d$ are positive integers with $ab=cd$, then there exist positive integers $p$, $q$, $r$, and $s$ such that $a=pq$, $b=rs$, $c=pr$, and $d=qs$.
Proof. We are given $ab=cd$. Set $p=\gcd(a,c)\ge1$, so that $a=pq$ and $c=pr$ for positive integers $q$ and $r$ with $\gcd(q,r)=1$. Hence $pqb = prd$, and dividing both sides by $p \ne 0$ leaves $qb = rd$. As $\gcd(q,r)=1$, the Fundamental Theorem of Arithmetic implies $r \mid b$, say $b=rs$ for an integer $s \ge 1$. Then $qrs=rd$, so $d=qs$.$\square$