I'm retaining as much as possible from your own wording; but note the differences!
The function $f(z):={z\over 1-|z|}$ is a function from the open unit disc $D$ to ${\mathbb C}$.
The function $f$ is onto if (not: "since") for every $w$ in the codomain ${\mathbb C}$, there exists a $z\in D$ such that $f(z)=w\ $, i.e., ${z\over 1-|z|}=w\ .\qquad(1)$ So by taking moduli: $|w|={|z|\over 1-|z| }$ or $|z|={|w|\over 1+|w|}\ .\qquad(2)$ On the other hand, taking arguments in $(1)$ for a $z$ with $|z|<1$ we get $\arg(z)=\arg(w)\ .\qquad(3)$ Equations $(2)$ and $(3)$ together imply that a $z$ of the required kind would necessarily be given by $z:={w\over 1+|w|}\ .$ Now we have arrived at this result not by means of a general theory about such problems, but by means of an ad-hoc procedure. Therefore we have to check whether the $z$ we have found indeed fulfills the conditions $z\in D$ and $f(z)=w$. I leave this verification to you.