The easiest way to deal with series like $\sum_{n=0}^\infty z^n w^{-n}$ is with iterated Laurent series. This series is an element of the ring $\mathbb{Z}((w))[[z]]$: power series in $z$ whose coefficients are Laurent series in $w$. (In this case Laurent polynomials in $w$ would suffice.)

A much more general, though more complicated, approach is through Hahn series (also called Mal'cev–Neumann series)
in which we have an indeterminate with exponents from an ordered group, with the condition that the exponents corresponding to nonzero terms are well ordered. (The well-ordered condition implies that multiplication of these series is well-defined.) To represent $\sum_{n=0}^\infty z^n w^{-n}$ in this way, we take as our exponent group the additive group $\mathbb{Z}\times\mathbb{Z}$ ordered lexicographically. With $x$ as the indeterminate, the series under consideration are of the form $\sum_{(i,j)\in \mathbb{Z}\times\mathbb{Z}} x^{(i,j)}$. We multiply monomials by $x^{(i_1,j_1)} x^{(i_2,j_2)}=x^{(i_1+i_2, j_1+j_2)}$. We may identify $x^{(i,j)}$ with $z^iw^j$. Then
\begin{equation*}
\sum_{n=0}^\infty z^n w^{-n}=\sum_{n=0}^\infty x^{(n,-n)}
\end{equation*}
is allowable since the exponent set $\{(0,0), (1,-1), (2,-2),\dots\}$ contains no infinite decreasing sequence. On the other hand
\begin{equation*}
\sum_{n=0}^\infty z^{-n} w^{n}=\sum_{n=0}^\infty x^{(-n,n)}
\end{equation*}
is not allowed since the exponent set contains the infinite decreasing sequence $(0,0)>(-1,1)>(-2,2)>\dots$.

5more comments