Let $X$ be a normed space, $\Omega\subseteq\mathbb C$ be open and $f:\Omega\to X$ be meromorphic in the sense that for all $z_0\in\Omega$, there is a $k_{z_0}\in\mathbb N_0$, $\left(a^{(z_0)}_k\right)_{k\ge-k_{z_0}}\subseteq X$ and a neighborhood $N_{z_0}\subseteq\Omega$ of $z_0$ with $$f(z)=\underbrace{\sum_{k=-k_{z_0}}^{-1}a_k^{(z_0)}(z-z_0)^k}_{=:\:m_{z_0}(z)}+\underbrace{\sum_{k=0}^\infty a_k^{(z_0)}(z-z_0)^k}_{=:\:h_{z_0}(z)}\tag1$$ for all $z\in N_{z_0}\setminus\{z_0\}$. Let $P$ denote the subset of $\Omega$ for which $k_{z_0}$ cannot be chosen to be $0$.
EDIT: The basic idea of this question is to consider a region $B$ of $f$, where $f$ has at most $|F|$ poles, and then to consider the Laurent expansion at the first pole, substract the principal part to obtain a function which has ony $|F|-1$ poles and continue this process until we arrive at a holomorphic function. This should somehow be similar to what is done in this answer, but I don't get why we arrive at a holomorphic function on all of $B$ at the end, since the "holomorphic parts" of each expansion are only defined on a small neighborhood of the corresponding pole ...
Let $B\subseteq\mathbb C$ be bounded and open with $\overline B\subseteq\Omega$. Are we able to show that for $F:=B\cap P$, we can decompose $f$ on $B$ into $$f(z)=\sum_{z_0\in F}\tilde m_{z_0}(z)+h_F(z)\;\;\;\text{for all }z\in B\tag2$$ for some holomorphic $h_F:B\to X$ and $$\tilde m_{z_0}(z)=\sum_{k=-k_{z_0}}^{-1}b_k^{(z_0)}(z-z_0)^k\;\;\;\text{for all }z\in\Omega$$ for some suitable $\left(b^{(z_0)}_k\right)_{k\ge-k_{z_0}}\subseteq X$?
Since $\overline B$ is compact and $P$ is discrete, $F$ must be finite and hence the sum over $F$ in $(2)$ is finite.
We may clearly assume that $N_{z_0}$ is open for all $z_0\in F$, if that would be an issue.
My idea is the following: Consider $z_i\in P$ with $z_1\ne z_2$. We know that $P$ is discrete and hence there is a $\varepsilon>0$ with $$|z_1-z_2|\ge\varepsilon\tag3.$$ Shrinking $N_{z_i}$ if necessary, we may assume that $$N_{z_i}\subseteq B_{\varepsilon/2}(z_i)\tag4$$ and hence $$N_{z_1}\cap N_{z_2}=\emptyset.\tag5$$ Let $$N_{\{z_1,\:z_2\}}:=N_{z_1}\cup N_{z_2}$$ and $$m_{\{z_1,\:z_2\}}:=\left.\begin{cases}m_{z_1}(z)&\text{, if }z\in N_{z_1}\\m_{z_2}(z)&\text{, if }z\in N_{z_2}\end{cases}\right\}\;\;\;\text{for }z\in N_{\{z_1,\:z_2\}}$$ (since the sum in the definition of $m_{z_i}$ is finite, $m_{z_i}$ is well-defined on all of $\Omega$). Shrinking $N_{z_i}$ if necessary, we may assume that $N_{z_1}$ and hence $N_{\{z_1,\:z_2\}}$ is open. If I'm not missing something, $$h_{\{z_1,\:z_2\}}(z):=\left.\begin{cases}h_{z_1}(z)&\text{, if }z\in N_{z_1}\\h_{z_2}(z)&\text{, if }z\in N_{z_2}\end{cases}\right\}\;\;\;\text{for }z\in N_{\{z_1,\:z_2\}}$$ should be holomorphic.
In the same way we should be able to construct a holomorphic function $h_F$ on $$N_F:=\bigcup_{z_0\in F}N_{z_0},$$ shrinking the $N_{z_0}$, if necessary, to ensure that this union is disjoint, with $$h_F(z)=h_{z_0}(z)\;\;\;\text{for all }z\in N_{z_0}\text{ and }z_0\in F\tag6.$$ On the other hand, by definition of $P$, the function $f$ should be holomorphic on the open set $B\setminus N_F$. Thus, the extension of $h_F$ to $B$, by setting it to $f$ on $B\setminus N_F$, should be holomorphic as well.
Now we may define $$m_F(z):=m_{z_0}(z)\;\;\;\text{for }z\in N_{z_0}\text{ and }z_0\in F.$$ However, while each individual $m_{z_0}$ can clearly be extended to all of $\Omega$, we don't arrive at the desired decomposition yet. How do we need to the choose $\left(b^{(z_0)}_k\right)_{k\ge-k_{z_0}}$?
$\frac{\pi^2}{\sin^2(\pi z)}$ is meromorphic,
$$\frac{\pi^2}{\sin^2(\pi z)}-\sum_{n=-N}^N \frac1{(z-n)^2}$$ is holomorphic on $|z|<N+1$ (complex differentiable at every point, a local property). The Cauchy integral formula implies that it is analytic and equal to its Taylor series at $0$ for $|z|<N+1$.
It works the same way when substracting finitely many poles to any meromorphic function.
When substracting infinitely many poles it gets more complicated, for example with $\frac{2i\pi z}{e^{2i\pi z}-1}$ then subtracting $\sum_n \frac{n}{z-n}$ won't work as it doesn't converge, instead we'll look at $\sum_n \frac{n}{z-n}+1+z/n$, generalizing this is the purpose of https://en.wikipedia.org/wiki/Mittag-Leffler%27s_theorem