Describing all holomorphic functions such that $f(n)=n$ for $n \in \mathbb{N}$ This question is inspired by a somewhat simpler one.

The question is: how can we classify all holomorphic functions $f:\mathbb{C}\rightarrow\mathbb{C}$ satisfying the property $\forall n \in \mathbb{N} \quad f(n)=n $?

If we have $g:\mathbb{C}\rightarrow\mathbb{C}$ such that $g\big|_\mathbb{N}\equiv 0$, then $f(z)=z+g(z)$ satisfies the criterion. Conversly, given such $f$ and defining $g(z)=f(z)-z$, we get $g\big|_\mathbb{N}\equiv 0$. So, the question boils down to classifying such $g$.

The set $I$ of such $g$, which is $I=\{g:\mathbb{C}\rightarrow\mathbb{C}, g\big|_\mathbb{N}\equiv 0\}$, is an ideal of the algebra of holomorphic functions, so we can ask for its generators. Obviously, $\forall k\in\mathbb{Z}\quad 1-e^{2\pi kz}\in I$, but I am not able to prove that they generate $I$.
Let $f$ any such function. Then $g(z)= f(z)/(ze^{2\pi iz})$ is entire and equal to $1$ at the natural numbers. 

Therefore $g(z)-1$ is zero at the naturals. Let $\Pi(z)$ be a Weierstrass product giving you an entire function vanishing exactly at the natural numbers.

Therefore all the $g$ are of the form $\Pi(z)h(z)$ for any entire $h(z)$.

Hence $g(z)=\Pi(z)h(z)+1$ and $$f(z)=ze^{2\pi i z}(\Pi(z)h(z)+1)$$

Two things change here to get all the functions. The product $\Pi(z)$ for all the orders that you may want the zeros to have, and the entire function $h$, which is arbitrary and you may just consider it part of $\Pi(z)$ anyway.
Let $f$ any such function. Then $g(z)= f(z)/(ze^{2\pi iz})$ is entire and equal to $1$ at the natural numbers. 

Therefore $g(z)-1$ is zero at the naturals. Let $\Pi(z)$ be a Weierstrass product giving you an entire function vanishing exactly at the natural numbers.

Therefore all the $g$ are of the form $\Pi(z)h(z)$ for any entire $h(z)$.

Hence $g(z)=\Pi(z)h(z)+1$ and $$f(z)=ze^{2\pi i z}(\Pi(z)h(z)+1)$$

Two things change here to get all the functions. The product $\Pi(z)$ for all the orders that you may want the zeros to have, and the entire function $h$, which is arbitrary and you may just consider it part of $\Pi(z)$ anyway.