A generalization of polynomial algebra on a Riemann surface

Question

Let $M$ be a $1$-dimensional complex manifold. Let $A$ be the space of all holomorphic functions $f:M\to \mathbb{C}$ such that either $f$ is a constant function or every level set $f^{-1}(c)$ is a finite (probably empty) set. Is $A$ an algebra of functions? Is its closure, with respect to topology of uniform convergence on compact subsets, equal to space of all holomorphic functions from $M$ to $\mathbb{C}$?

Answer 1

Counterexample to the first question ("is $A$ an algebra of functions?"): Let $M$ be a vertical strip such as {$x + iy : 0 < x < 1$}, and define $f_1,f_2$ as the restriction to $M$ of the entire functions $$ f_1(z) = \exp((1+i)z), \quad f_2(z) = \exp((1-i)z). $$ Then $f_1,f_2 \in A$ but $f_1 f_2 \notin A$. Indeed if $f_1(z) = f_1(z')$ or $f_2(z) = f_2(z')$ then the real part of $z-z'$ is a multiple of $\pi$, so all the level sets of $f_1$ and $f_2$ are of bounded size (indeed size at most $1$ for our width-$1$ strip). But $f_1 f_2$ is the function $z \mapsto e^{2z}$, whose level sets are translates of $\pi i \bf Z$ and thus can contain infinitely many points of $M$.

It follows that $A$ is not closed under addition either: if it were, then $A$ would contain $f_1+f_2$ and $f_1-f_2$, and since $A$ is closed under squaring this would imply that $A \ni (f_1 + f_2)^2 - (f_1 - f_2)^2 = 4 f_1 f_2$.