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 $c_k = \sum_l f_l g_{k-l}$.
To evaluate the integral, I'd proceed this way. The integrand equals
$$\frac{\sin z}{(z+3)^2}\cdot \frac{1}{(z-1)^2}.$$
We know that first quotient, call it $f,$ is analytic at $1.$ Hence it will have a Taylor series there that looks like $f(z) = f(1) +f'(1)(z-1) + \cdots.$ Thus our integrand, near $1,$ equals
$$\frac{f(1)}{(z-1)^2} + \frac{f'(1)}{z-1} + \cdots.$$
It follows that the residue of the integrand is $f'(1),$ and the integral equals $2\pi i\cdot f'(1).$ Calculate $f'(1)$ and you're done.
(I don't understand some of the questions you raise. Perhaps if you asked a specific question about what I did ...?)
Best Answer
Let $r \in (0,1)$ and let $n \in \mathbb Z$ such that $n \le -1$. We have
$a_n= \frac{1}{2 \pi i} \int_{|z|=r} \frac{f(w)}{w^{n+1}} dw$.
An easy estimation gives
$|a_n| \le r^{-n-1/2} \sin(1/r)$.
Since $r \in (0,1)$ was arbitrary and $\sin(1/r)$ is bounded, we get with $r \to 0$ that $a_n=0$.