The idea is that after subtracting off the pole at $z=z_0$, you are left with something holomorphic. After expanding in a Laurent series around $z=z_0$ as you suggest, and subtracting off $P(z):=\sum_{k>0} A_k (z-z_0)^{-k}$ from $f(z)$, you are left with something that is holomorphic at $z=z_0$ (since you removed the singular part) and also holomorphic in the remainder of the region $\{z\mid |z|<R\}$ (since both $f(z)$ and $P(z)$ are holomorphic in $\{z\mid z\ne z_0,\ |z|<R\}$). Therefore, $g(z)=f(z)-P(z)$ is holomorphic on $\{z\mid |z|<R\}$, so it has a power series around $0$ with radius of convergence at least $R$.
I'm going to take $z_0=0$ for simplicity. For $\rho_1, \rho_2$ with $r_1 < \rho_1 < \rho_2 < r_2$, and for $z$ with $\rho_1 < |z| < \rho_2$, Cauchy's integral formula applied to the annulus $\{ \rho_1 \leq |w| \leq \rho_2\}$ gives us that
$$f(z)= \frac{1}{2\pi i} \int_{|w|=\rho_2} \frac{f(w)}{w-z} \ \mathrm{d}w - \frac{1}{2 \pi i} \int_{|w|=\rho_1} \frac{f(w)}{w-z} \ \mathrm{d}w.$$
In the first integral we can write
$$\frac{f(w)}{w-z} = \sum_{n=0}^\infty \frac{z^nf(w)}{w^{n+1}},$$
and for each $z$ with $|z| < \rho_2$ this sum converges aboslutely uniformly on the circle $\{|w|=\rho_2\}$.
[Added in response to comment: there exists a positive real number $M$ such that $|f(w)| \leq M$ on $\{|w|=\rho_2\}$. Then for all $w, z$ with $|w|=\rho_2$ and $|z|<\rho_2$, and all $n \geq 0$, we have
$$\Bigg|\frac{z^nf(w)}{w^{n+1}}\Bigg| \leq \frac{M}{\rho_2} \Bigg(\frac{|z|}{\rho_2}\Bigg)^n.$$
The right-hand side converges (it's a geometric series with ratio $|z|/\rho_2 <1$) so the claimed sum converges absolutely uniformly on the circle by the Weierstrass M-test.]
So the first integral is equal to
$$\sum_{n=0}^\infty z^n \frac{1}{2\pi i} \int_{|w|=\rho_2} \frac{f(w)}{w^{n+1}} \ \mathrm{d}w,$$
and this converges and is valid on the whole of $\{|z| < \rho_2\}$. Letting $\rho_2 \rightarrow r_2$, we see that the positive part of the Laurent series converges on $\{|z| < r_2\}$.
For the negative part, do a similar expansion with $w^n/z^{n+1}$.
Best Answer
There is a transformation error in the question. Instead of $$f(z) = \sum_{n \geq 0} a_n (z-z_0)^n \sum_{k=0}^n {n \choose k} (z-z_0)^{-k} z_0^k = \sum_{n \geq 0} c_n (z-z_0)^n,$$ with $c_n = a_n \sum_{k=0}^n (z-z_0)^{-k} z_0^k$, it should have been
$$f(z)=\sum_{n=0}^\infty a_nz^n=\sum_{n=0}^\infty c_n(z-z_0)^n,$$ where $c_n=\sum_{s=n}^\infty a_s\binom{s}{n} z_0^{s-n}$
More explicitly, as a formal computation,
$$\begin{align} f(z)&=\sum_{n=0}^\infty a_nz^n= \displaystyle\sum_{n=0}^\infty a_n(z_0+(z-z_0))^n\\ &=\sum_{n=0}^\infty\sum_{k=0}^n a_n\binom{n}{k} z_0^{n-k}(z-z_0)^{k}\\ (\text{Switching the order of summation})\quad &=\sum_{k=0}^\infty\sum_{n=k}^\infty a_n\binom{n}{k} z_0^{n-k}(z-z_0)^{k}\\ (\text{Replace $n$ by $s$ and $k$ by $n$ simultaneously})\quad &=\sum_{n=0}^\infty\left(\sum_{s=n}^\infty a_s\binom{s}{n} z_0^{s-n}\right)(z-z_0)^{n}.\\ \end{align}$$
If we substitute a complex number near $z_0$ for $z$ in the formal computation above, do all infinite sums thereof converge? Do all equalities hold? This is the question asked.
The answer is yes.
The basic idea is that absolute convergence implies convergence and, furthemore, the order of summation does not matter. That general principle can be stated in very general ways. What we need here is the following specific manifestation.
If $d(k,n)\in \mathbb C$ for all integer $0\le k\le n$ and $\sum_{n=0}^\infty\sum_{k=0}^n \lVert d(k,n)\rVert$ converges, then we have the following.
(If you are not familiar with absolute convergence, it should be a good exercise to prove the above.)
All we need to check now is the convergence of
$$\sum_{n=0}^\infty\sum_{k=0}^n \left\lVert a_n\binom{n}{k} z_0^{n-k}(z-z_0)^{k}\right\rVert.$$
Here we go. Suppose $\rVert z-z_0\rVert\lt R-\lVert z_0\rVert$. Then $\lVert z_0\rVert+ \lVert(z-z_0)\rVert\lt R$. We have
$$\begin{align} \sum_{n=0}^\infty\sum_{k=0}^n \left\lVert a_n\binom{n}{k} z_0^{n-k}(z-z_0)^{k}\right\rVert &= \sum_{n=0}^\infty\sum_{k=0}^n \lVert a_n\rVert\binom{n}{k} \lVert z_0\rVert^{n-k}\lVert(z-z_0)\rVert^k \\ &= \sum_{n=0}^\infty\lVert a_n\rVert(\lVert z_0\rVert+ \lVert(z-z_0)\rVert)^n\\ \end{align}$$
Since $R$ is the radius of convergence of $\sum_{n=0}^\infty a_nz^n$, the right-hand size converges. $\blacksquare$