I want to know if there exist a known expansion or can be derived of the polynomial $$ \prod_{i=1}^{m}\prod_{j= 1}^{n}(1-z(x_i + y_i))^{-w} \tag{*}$$ in terms of Schur function. That is asking for (*) can be written in the from $$ \sum_{\lambda , \nu} d_{\lambda , \nu} s_{\lambda}({\bf{x}})s_{\mu}({\bf{y}}) $$ what is this $d_{\lambda , \nu}$ ? $\lambda$, $\mu$ are ranging over all partitions.
Let me give you things that are close to the expression known in literature and can be found in I.G Macdonald Symmetric function and Hall polynomials.
$$\prod_{i=1}^{m}(1-x_i)^{-n} = \sum_{\lambda, \ell({\lambda})\leq m} d_{\lambda}(n)s_{\lambda} \,\,\, (n\in \mathbb{Z}_{\geq 1})\tag{**}$$ where $$d_{\lambda}(n) = \prod_{(i,j)\in \lambda}\frac{n+j-i}{h_{\lambda}(i,j)} $$, $\ell(\lambda)$ denote the length of $\lambda$, $h_{\lambda}(i,j)$ is the hook length.
If $m=1$ we can see that that imply that the length of the partition lambda can be only 1, hence $\lambda$ can be of the from $(k)$, where $k$ is a positive integer. hence $$(1-x_1)^{-n} = \sum_{k\geq 1}\prod_{i=1}^{k}\frac{(n + i-1)}{i} s_{k}(x_1)z^n $$
The other identity is given by
$$ \prod_{i=1}^{m}\prod_{j= 1}^{n}(1+(x_i + y_i))= \sum_{{\mu}\subset \lambda \subset (n^m)}d_{\lambda , \nu} s_{\lambda^{'}}({\bf{x}})s_{\mu}({\bf{y}})\tag{*}$$ $\lambda^{'}$ represent the conjugate partition to lambda.
In $(**)$ if we replace $x_i$ by $x_i +y_1$ we can get an expression on the RHS involving $s_{\lambda}(x_1+ y_1, x_2+y_1,\ldots)$. Now we can do it all $y_j$ and get such expression. I guess then the question reduces to find $$s_{\lambda}(x_1 + y_1 , x_2 +y_1, \ldots) $$ can be expressed in terms of Schur functions ?
