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 $c_k = \sum_l f_l g_{k-l}$.
The Laurent series in any given annulus is uniquely determined. (You should look up a proof of this. It's important.) In this case, you have convergence in a punctured disk. To determine this disk, look at the singularities of the function and note the distance from 1 to the nearest singularity. (If you return to the proof of the existence of Laurent series, you will see that the radius of convergence will be the largest possible punctured disk around with point that doesn't contain another singularity.)
The hard part about finding the Laurent series here is inverting the denominator. If you write out the power series for the denominator, you will see that it is
$$(z-1)^2(a+b(z-1)+c(z-1)^2+\dots)$$
where $a,b,c$ are some constants. So you can factor out the $\frac{1}{(z-1)^2}$ term and use the normal procedure for inverting power series with a nonzero constant term. You can use your idea (writing everything in terms of $z-1$) to take care of the numerator.
Best Answer
$$\frac1{z-1}=\frac1{z-3+2}=\frac12\frac1{1+\frac{z-3}2}=\frac12\left(1-\frac{z-3}2+\frac{(z-3)^2}4-\ldots\right)=$$
$$=\frac12-\frac{z-3}4+\ldots$$
So
$$\frac z{(z-1)(z-3)}=\frac12\left(\frac3{z-3}-\frac1{z-1}\right)=\frac32\frac1{z-3}-\frac14+\frac{z-3}8+\ldots$$
so the residue is $\,3/2\,$ .