This identity is taken from a physics paper [1] stated without proof, on page 43.
$$ \frac{1}{(q)_\infty} \sum_{m \in \mathbb{Z}} q^{\frac{m^2}{2}} (-q^{-\frac{1}{2}}x)^m y^m(q^{1-m}y^{-1};q)_\infty = \frac{(xy;q)_\infty(qx^{-1}y^{-1};q)_\infty}{(x;q)_\infty} \tag{$*$}$$
This formula is new and interesting and probably correct. I don't know a good starting point. While this is a $q$-series, there are two deformation paramters $x$ and $y$. It may help to recall some definitions:
$\displaystyle \prod_{n \geq 0} \big( 1 - q^n x \big) =: (x;q)_\infty $ this is the $q$-Pochhammer symbol
$\displaystyle \prod_{n = 0}^{k-1} \big(1 - q^n x \big) =: (x;q)_k$ this is the $q$-Pochhammer symbol
The symbol $(q;q)_k$ gets its own abbreviation $(q)_k$ but it's very similar.
$\displaystyle (q)_\infty := (q;q)_\infty = \prod_{n \geq 1} (1 - q^n) $
$\displaystyle (q)_k := (q;q)_k = \prod_{n=1}^k (1 - q^n) $
These series are likely to "converge" as elements of $\mathbb{Z}[q^{\frac{1}{2}}][[x^{\pm 1}, y^{\pm 1}]]$, but do these also converge analytically? This might not converge for any $q \in \mathbb{C}$.
In order to check the equivalence, we might start with some partial fraction decomposition:
$$ \frac{1}{(x;q)_\infty} = \sum_{n=0}^\infty \frac{x^n}{(q;q)_n} $$
I also wonder what the RHS of $(*)$ is counting.