From Terrence Tao's Blog:
I don't see how this define an equivalence. $f_1 $ ~ $f_2$ , $f_2$ ~ $f_3$ implies $f_1$ and $f_3$ coincide on $U\setminus (S_1 \cup S_2 \cup S_3)$ and not $U\setminus (S_1 \cup S_3)$? This doesn't satisfy definition of equivalence?
(By the way, how do I rescale the image?)
As you've pointed out, it turns out that this fact is not completely trivial. It relies on the fact that meromorphic functions are continuous and are defined everywhere except on a discrete set. If $f_1$ and $f_2$ agree in this sense and $f_2$ and $f_3$ agree, it seems like there may be a point $z$ where $f_2$ is not defined but $f_1$ and $f_3$ are, and $f_1(z)\neq f_3(z)$. But this is not possible because $$f_1(z)=\lim_{z'\to z}{f_1(z')}=\lim_{z'\to z}{f_3(z')}=f_3(z)$$ The equality in the middle is true because $f_1$ and $f_3$ agree on every point in a punctured neighborhood of $z$.