Your reasoning is pretty much correct. Since $f$ is holomorphic on $D_r(z_0)\setminus\{z_0\}$, and $\lim_{z\to z_0}f(z)=\frac{1}{g(z_0)}-w$ exists, it follows the singularity at $z_0$ is removable, contradicting that it is essential. This is one of the many ways to tell if a singularity is removable.
This is what they mean when they say $f$ is holomorphic at $z_0$; they mean the singularity is removable there, so it can be made holomorphic by an appropriate definition of $f(z_0)$.
I am writing to add some necessary details to the proof of the statement $\textbf{2}$.
The OP didn't provide his definition of $\textit{pole}$. By inspecting his proof, I guess his definition of $\textit{pole}$ is based on the singular part of Laurent series. To be more specific, his definition might be that, if the Laurent series at a singularity $z_0$ has at least one and at most finitely many terms of negative powers, then $z_0$ is called a $\textit{pole}$.
So in the proof of the converse direction of the statement $\textbf{2}$, we have to show that the Laurent series of $f$ at $z_0$ has at least one and at most finitely many terms of negative powers.
Proof:
First let's suppose that $\text{lim}_{z\to z_0}f(z)=\infty$.
Let $g(z):=\frac{1}{f(z)}$. Then $g(z)$ is well-defined in a ball centered at $z_0$ and $\text{lim}_{z\to z_0}g(z)=0.$ This implies that $g$ is holomorphic in the ball.
Since $g$ is holomorphic in the ball and has a zero at $z_0$, we can find some holomorphic function $h$ s.t. $g(z)=(z-z_0)^k h(z)$ with $h(z_0)\neq0$, where $k\ge 1.$
Since $h$ is holomorphic in the ball and $h(z_0)\neq0$, we may assume $h$ is nonzero in the ball. (Because we can always shrink the ball to a smaller ball, it doesn't really matter.) So $\frac{1}{h}$ is holomorphic in the ball and hence we can write $\frac{1}{h(z)}=\sum_{n=0}^\infty a_n(z-z_0)^n$ for some coefficients $\{a_n\}$ by the Taylor theorem.
So at every point $z$ in the punctual ball (i.e. the ball excluding $z_0$), we have $$f(z)=\frac{1}{g(z)}=\frac{1}{(z-z_0)^k h(z)}=\sum_{n=0}^\infty a_n(z-z_0)^{n-k}=\sum_{n=-k}^\infty a_{n+k}(z-z_0)^n.$$
By the definition, we conclude that $z_0$ is indeed a pole.
$$\tag*{$\blacksquare$}$$
Best Answer
Suppose that $\lim_{w\to z}f(w)=l(\in\mathbb{C})$. Take $\varepsilon>0$. There is the a $\delta>0$ such that $\lvert w-z\rvert<\delta\wedge w\neq z\implies\bigl\lvert f(w)-l\bigr\rvert<1$. That is,$$f\bigl(D'(z,\delta)\bigr)\subset D(l,1).$$But then$$\overline{f\bigl(D'(z,\delta)\bigr)}\subset\overline{D(l,1)}\varsubsetneq\mathbb C.$$Can we deal with case in which $f$ has a pole on $z$ now?