You can obtain the desired result using absolute convergence and some observations about the annulus of convergence of Laurent series.
If $\{x_n\}_{n \in \mathbb{Z}}$, then let $R_+(\{x_n\}) = ( \limsup_{n \to \infty} \sqrt[n]{|x_n|} )^{-1}$, and $R_-(\{x_n\}) = \limsup_{n \to \infty} \sqrt[n]{|x_{-n}|}$.
Two relevant results (all summations are over $\mathbb{Z}$):
(i) If $\sum_m \sum_n |a_{m,n}| < \infty$, then the summation can be rearranged. In particular, $\sum_m \sum_n a_{m,n} = \sum_k \sum_l a_{l,k-l}$. (Since $\phi(m,n) = (m,m+n)$ is a bijection of $\mathbb{Z}^2$ to $\mathbb{Z}^2$.)
(ii) If $\sum_{n} |x_n| < \infty$, then $R_+(\{x_n\}) \ge 1$. Similarly, $R_-(\{x_n\}) \le 1$.
Let $f(z) = \sum_n f_n (z-z_0)^n$, $g(z) = \sum_n g_n (z-z_0)^n$. Let $R_+ = \min(R_+(\{f_n\}),R_+(\{g_n\}))$, $R_- = \max(R_-(\{f_n\}),R_-(\{g_n\}))$.
Choose $R_-<r < R_+$. Then $\sum_m |f_m| r^m$ and
$\sum_n |g_n| r^n$ are absolutely convergent sequences, and so $\sum_m \sum_n |f_m||g_n| r^{m+n} < \infty$. From (i) we have $\sum_m \sum_n |f_m||g_n| r^{m+n} = \sum_k (\sum_l |f_l| |g_{k-l}|) r^k < \infty$.
If we let $c_k = \sum_l f_l g_{k-l}$, this gives $\sum_k |c_k| r^k < \infty$, and so
$R_+(\{c_kr^k\}) = \frac{1}{r} R_+(\{c_k \}) \ge 1$, or, in other words, $R_+(\{c_k \}) \ge r$. Since $r<R_+$ was arbitrary, it follows that $R_+(\{c_k \}) \ge R_+$. The same line of argument gives $R_-(\{c_k \}) \le R_-$.
Hence we have $c(z) = f(z)g(z) = \sum_n c_n (z-z_0)^k$ on $R_- < |z-z_0| < R_+$, where $c_k = \sum_l f_l g_{k-l}$.
I wanted to prove the convergence for both parts of the series, because if both parts converge, then also the whole Laurent series.
This is a very good idea, indeed:
The power series $\sum_{n=0}^\infty a_n(z-z_0)^n$ converges in the disk $|z-z_0|<R$ where $R$ is given by the Cauchy-Hadamard formula $$R^{-1}= \limsup |a_n|^{1/n} $$
The series of negative terms can be transformed into a power series with positive powers using the substitution $\frac{1}{z-z_0}=w$. The series then becomes $$\sum_{n=1}^\infty a_{-n} w^n $$ and it converges in the disk $|w|<r$, where, again by Cauchy-Hadamard's formula $r^{-1}=\limsup |a_{-n}|^{1/n}$. Clearly the disk $|w|<r$ is equivalent to the annulus $\left| \frac{1}{z-z_0}\right|<r$.
Combining the above results, the problem is solved.
Best Answer
Remember that a Laurent series is just the sum of two power series, one in $u = (z - z_0)$ and the other in $w = \dfrac 1{z-z_0}$. Each power series has a radius of convergence, say $r_u$ for the series in $u$ and $r_w$ for the series in $w$. Power series are absolutely convergent everywhere inside their radius of convergence (if you look at the proof of Hadamard's formula for the radius of convergence, you will see that it actually proves absolute convergence).
But $$|u| < r_u \implies|z-z_0| < r_u$$
and $$|w| < r_w \implies \dfrac 1{|z-z_0|} < r_w \implies \dfrac 1{r_w} < |z - z_0|$$
So setting $r_2 = r_u$ and $r_1 = \frac 1{r_w}$ gives that the Laurent series is absolutely convergent on $r_1 \le |z - z_0| \le r_2$.