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 $c_k = \sum_l f_l g_{k-l}$.
Suppose there's another expansion
$$f(z)=\sum_{-\infty}^\infty b_n z^n.$$
But then we would have
$$0=f(z)-f(z)=\sum_{-\infty}^\infty a_n z^n-\sum_{-\infty}^\infty b_n z^n=\sum_{-\infty}^\infty (a_n-b_n) z^n.$$
But for this to be true $\forall z$ in the domain, we must have $a_n-b_n=0, \forall n$. So, the coefficients, are unique.
Best Answer
One way to go is to first note that $$z^2+z^4=z^2\left(z^2+1\right)=z^2(z-i)(z+i),$$ and that the numerator's degree is less than that of the denominator. Thus, there are constants $a,b,c,d$ such that $$\frac{2+3z}{z^2+z^4}=\frac{a}{z}+\frac{b}{z^2}+\frac{c}{z-i}+\frac{d}{z+i}.$$ I leave it to you to find them. (This is known as a partial fraction decomposition, if you want to look up more about it.)
Next, we could use the fact that for $z\ne 0,$ we have that $$\frac{1}{z+i}=\frac{1}{z}\cdot\frac{1}{1-\left(-\frac{i}{z}\right)}$$ and that $$\frac{1}{z+i}=\frac{1}{i}\cdot\frac{1}{1-\left(-\frac{z}{i}\right)},$$ then use the fact that for $|w|<1,$ we know that $$\frac{1}{1-w}=\sum_{n=0}^\infty w^n.\tag{$\heartsuit$}$$ In order to make $|z|<1,$ it is enough to make $|w|<1$ for one of $w=-\frac{i}{z}$ or $w=-\frac{z}{i}.$ (I'll leave it to you to determine which one.) That will let us rewrite $\frac{d}{z+i}$ as a Laurent series that is valid in the annulus $0<|z|<1,$ and we can similarly rewrite $\frac{c}{z-i}.$ Since $\frac{a}{z}+\frac{b}{z^2}$ is already a Laurent series that's valid in the given annulus, then we can simply add them together, and obtain the series $$\sum_{n=0}^\infty a_nz^n+\sum_{n=1}^\infty b_nz^{-n}$$ that we wanted. Note the slight difference between mine and yours. Can you see the reason for it?
The above approach is probably good for you to run through, for practice's sake. However, in this case, we can proceed in an even easier way! Note that $$f(z)=\frac{2+3z}{z^2}\cdot\frac{1}{1-\left(-z^2\right)},$$ and that $0<|z|<1$ if and only if $0<\left|z^2\right|<1,$ so we can use $(\heartsuit)$ to conclude that $$\frac{1}{1-\left(-z^2\right)}=\sum_{n=0}^\infty\left(-z^2\right)^n=\sum_{n=0}^\infty(-1)^nz^{2n}.$$ Hence, $$f(z)=\sum_{n=0}\frac{2+3z}{z^2}\cdot(-1)^nz^{2n}=\sum_{n=0}^\infty(-1)^n(2+3z)z^{2n-2}=\sum_{n=0}^\infty(-1)^n2z^{2n-2}+(-1)^n3z^{2n-1}.$$ Can you take it from there to figure out what the sequences $\left\{a_n\right\}_{n=0}^\infty$ and $\left\{b_n\right\}_{n=1}^\infty$ should be?