I keep the notation $(q)_s=\prod_{i=1}^s(1-q^i)$, and in addition set `\begin{equation}\binom{a}{s}_q=\frac{(q)_a}{(q)_s(q)_{a-s}},\end{equation}`

the Gaussian binomial coefficient.

Comparing coefficients, the question amounts to showing that
```
\begin{equation}
\sum_s\binom{a}{s}_q\binom{b}{s}_q(q)_sq^{(a-s)(b-s)}=1.
\end{equation}
```

This, however follows immediately from induction over $a$: The assertion is obvious for $a=0$. Now suppose $a\ge1$. Use
```
\begin{equation}
\binom{a}{s}_q=q^s\binom{a-1}{s}_q+\binom{a-1}{s-1}_q
\end{equation}
```

to replace $\binom{a}{s}_q$. The contribution from the first summand of the right hand side gives $q^b$ (I omit the trivial calculation), by applying the induction hypothesis to $a-1$. In order to handle the contribution from the second summand, we note that
```
\begin{equation}
\binom{b}{s}_q(q)_s=(1-q^b)(q)_{s-1}\binom{b-1}{s-1}_q.
\end{equation}
```

So, applying the induction hypothesis for $a-1$ once again, we see that this other contribution is $1-q^b$. Now $q^b+(1-q^b)=1$ yields the result.