Write formula in elementary symmetric polynomials Consider the expression
$$
\prod_{i\in I, j\in J} (x-\alpha_i - \beta_j)
$$
as polynomial in $x$. As this expression is symmetric in the $\alpha_i$ and in the $\beta_j$, you should be able to write it in the elementary symmetric polynomials of $\alpha_i$ and $\beta_j$. 
I was able to do so for small examples, but I am looking for general formula or algorithm to derive the polynomial in $x$ with only elementary symmetric functions of $\alpha_i$ and $\beta_j$ as coefficients.
 A: This is just a partial answer and a sketch at that, but too big for a comment.

Let $I=\{1,\ldots,m\}$ and $J=\{1,\ldots,n\}$ so that
$$\prod_{i\in I, j\in J} (x-\alpha_i - \beta_j)
=\prod_{i=1}^m\prod_{j=1}^n((x-\alpha_i)-\beta_j)
=\prod_{i=1}^m\sum_{j=0}^ne_{n-j}(\beta_1,\ldots,\beta_n)(x-\alpha_i)^j,$$
and with the binomial theorem this becomes
$$\prod_{i=1}^m
\sum_{j=0}^ne_{n-j}(\beta_1,\ldots,\beta_n)
\sum_{k=0}^j(-\alpha_i)^k\binom{j}{k}x^{j-k}
=\prod_{i=1}^m
\sum_{j=0}^n\left(e_{n-j}(\beta_1,\ldots,\beta_n)
\sum_{k=0}^{n-j}(-\alpha_i)^k\binom{j+k}{k}\right)x^j.$$
For the sake of legibility let 
$$f_{n-j}(\alpha_i):=\sum_{k=0}^{n-j}(-\alpha_i)^k\binom{j+k}{k}
\qquad\text{ and }\qquad
e_{n-j}(\beta):=e_{n-j}(\beta_1,\ldots,\beta_n).$$
Then expanding the product above yields
\begin{eqnarray*}
\sum_{\ell=0}^{mn}\left(
\prod_{\substack{0\leq l_i\leq n\\\sum_il_i=\ell\\\sum_i1=m}}
e_{l_i}(\beta)f_{l_i}(\alpha_i)\right)x^{\ell}
&=&\sum_{\ell=0}^{mn}\left(\prod_{k=1}^n
(e_k(\beta)f_k(\alpha_i))^{\binom{(n-k)+(m-2)}{m-2}}\right)x^{\ell}\\
&=&\sum_{\ell=0}^{mn}\left(\prod_{k=1}^ne_k(\beta)^{\binom{(n-k)+(m-2)}{m-2}}\right)\left(\prod_{k=1}^n
f_k(\alpha_i)^{\binom{(n-k)+(m-2)}{m-2}}\right)x^{\ell},
\end{eqnarray*}
and then I'm tempted to say that by symmetry (in $\alpha$ and $\beta$) we must have
$$\prod_{k=1}^n
f_k(\alpha_i)^{\binom{(n-k)+(m-2)}{m-2}}=\prod_{k=1}^me_k(\alpha),$$
but this seems unlikely.
