Can a meromorphic function have removable singularities?

Question

Authors usually give the following definition (or some variation) of meromorphic function:

Let $G$ be a region (i.e. open connected subset of $\mathbb{C}$). A function $f$ is said to be meromorphic in $G$ if it's holomorphic in $G$ except in a discrete set of poles.

However I think we should also consider removable singularities in the definition of meromorphic function. There is a theorem that says:

If $f$ and $g$ are holomorphic in $G$ then $f/g$ is meromorphic in $G$.

This theorem is no longer true if we take the usual definition. Take for example $f,g:\mathbb{C}\to \mathbb{C}, f(z)=g(z)=z$ and define $h:\mathbb{C}-\{0\}\to\mathbb{C}, h(z)=f(z)/g(z)$. Then $h=1$ in $\mathbb{C}-\{0\}$, however $h$ is not meromorphic in $\mathbb{C}$ in the sense of the previous definition because $0$ is not a pole, is a removable singularity.

Most authors only mention poles in the definition of meromorphic functions, but I think that's not strictly correct. If I'm right, I would like to see an author considering removable singularities in the definition of meromorphic functions.

There is a similar question here definition of meromorphic function : it may have removable singularities but I don't like the phrasing of the question and the answer.

I would like to have a reference text considering these details. For example Lang, Conway and Ahlfors all give the definition without removable singularities.

Edit: I wrote an answer in the other question that is correct but I don't think it addresses the most important points. Below I will write a proper answer

I think the book Lectures on Riemann surfaces by Otto Forster points to the right direction. The most important thing is that for a (connected) Riemann Surface $X$, the set $\mathcal{M}(X)$ of meromorphic functions on $X$ should be a field and to do that, one sometimes has to extend the domain of definition of a given function. I'm just gonna copy the relevant part of the book here.

1.12. Definition. Let $X$ be a Riemann surface and $Y$ be an open subset of $X$. By a meromorphic function on $Y$ we mean a holomorphic function $f: Y' \to \mathbb{C}$, where $Y' \subset Y$ is an open subset, such that the following hold:

1. $Y \setminus Y'$ contains only isolated points.
2. For every point $p \in Y \setminus Y'$, one has \begin{equation*} \lim_{x \to p}|f(x)| = \infty \end{equation*} The points of $Y \setminus Y'$ are called the poles of $f$. The set of all meromorphic functions on $Y$ is denoted by $\mathcal{M}(Y)$.

1.13. Remarks

1. Let $(U, z)$ be a coordinate neighborhood of a pole $p$ of $f$ with $z(p) = 0$. Then $f$ may be expanded in a Laurent series \begin{equation*} f = \sum_{v = -k}^\infty c_vz^v \end{equation*} in a neighborhood of $p$.
2. $\mathcal{M}(Y)$ has the natural structure of a $\mathbb{C}$-algebra. First of all the sum and the product of two meromorphic functions $f, g \in \mathcal{M}(Y)$ are holomorphic functions at those points where both $f$ and $g$ are holomorphic. Then one holomorphically extends, using Riemann's Removable Singularities Theorem, $f + y$ (resp. $fg)$) across any singularities which are removable.

1.16. Remark. From (1.11) and (1.15) it follows that the Identity Theorem also holds for meromorphic functions on a Riemann surface. Thus any function $f \in \mathcal{M}(X)$ which is not identically zero has only isolated zeroes. This implies that $\mathcal{M}(X)$ is a field.