We turn our attention to arbitrary meromorphic binary operations $B(x,y): \mathbb{C}^2 \rightarrow \mathbb{C}$
I was considering so called "homomorphism" functional equations of the form $$ F(B(x,y)) = B(F(x), F(y)) $$
Some simple examples include
$$ F(x-y) = F(x) - F(y) $$
Which has meromorphic solutions $F(x) = c*x$ for your choice of complex constant $c$.
$$ F(x/y) = F(x)/F(y) $$
Which has mermorphic solutions $F(x) = x^c$ for your choice of complex constant $c$.
Now here's where things tricky. Suppose we look into
$$ F\left( \frac{1}{a-b} \right) = \frac{1}{F(a) - F(b)} $$
By setting $a = y + h$, $b = h$ its easy to see then that
$$ \lim_{h \rightarrow 0} \ h F \left( \frac{1}{h} \right) = \frac{1}{F'(y)} $$
Regardless whether the left hand side is well defined or not it clearly does not depend on $y$ at all, so any such $F$ must be piece wise constant and outside of the trivial solution $F(x) = x$ cannot be meromorphic.
So here's the question:
Under WHAT conditions on the binary operator $B(x,y)$ does the equation
$$ F(B(x,y)) = B(F(x), F(y)) $$
Have a non trivial meromorphic solution $F(x)$?
Possible Direction:
It's interesting that while $F \left( \frac{1}{a-b} \right) = \frac{1}{F(a) - F(b)}$ doesn't admit non trivial solutions that the equation $$ F \left( \frac{1}{\frac{1}{a} - \frac{1}{b}} \right) = \frac{1}{\frac{1}{F(a)} - \frac{1}{F(b)}} $$ DOES admit non trivial solutions, namely $f(x) = cx$
It would seem to me based on all this evidence that $B(x,y)$ having an identity and being associative might be very important.