$S_k(x+y)-S_k(x)-S_k(y)$ where $S_k$ is symmetric polynomial Let $S_k$ be the $k$-th elementary symmetric polynomial of $n$ variables. How can I rewrite $$S_k(x+y)-S_k(x)-S_k(y)$$ by just using $x,y,S_1,S_2,\cdots S_{k-1}$ where $x=(x_1,x_2,\cdots,x_n)$ and $y=(y_1,y_2,\cdots,y_n)$.
Example: Let $x=(x_1,x_2)$ and $y=(y_1,y_2)$
$$S_2(x+y)-S_2(x)-S_2(y)=(x_1+y_1)(x_2+y_2)-x_1x_2-y_1y_2$$
$$=x_1y_2+x_2y_1=(x_1+x_2)(y_1+y_2)-(x_1y_1+x_2y_2)$$
$$=S_1(x)S_1(y)-S_1(xy).$$
How can we generalize this for any $n$ and $k$?
I believe somebody found this before but my research area is far to symmetric polynomials. References are also accepted. Thanks.
Edit: (or addition:) It can be closed form or an algorithm.. Useful comments.
 A: Express $S_k$ in power sums via the Newton formula. For power sums this is the binomial formula for each summand. Then express power sums again in elementary symmetric functions.
Added:
The expression $S_k(x+y)-S_k(x)-S_k(y)$ is still invariant under the diagonal action of the symmetric group $\mathfrak S(n)$ acting on $x$ and $y$ in the same way. So it can written as a polynomial in a basis for this diagonal representation. One has to determine such a basis.
By Corollary 2.17 of here the algebra of invariant polynomials of the diagonal action is the integral closure of the the algebra of 2-polarizations of the algebra of symmetric polynomials. By Example 2.18, in the case of the permutation group, the algebra of 2-polizations is already integrally closed. 
However, the expression $S_2(x+y)-S_2(x)-S_2(y)$ is a 2-polarization. 
A: Not quite an answer. The logarithmic derivative of the generating function trick (as described very well in Yakovlenko's notes would seem to give a reasonable approach to this (I am not quite up to working through it right this moment).
A: This is just a suggestion how to proceed in the case $k=3$, which is too long for a comment.
In case of degree $=$ number of variables $=3$ we have this formula:
\begin{align}
 & S_3(x+y+z)-S_3(x+y)-S_3(y+z)-S_3(z+y)+S_3(x)+S_3(y)+S_3(z) \\
=\;& S_1(x)S_1(y)S_1(z)-S_1(xy)S_1(z)-S_1(yz)S_1(x)-S_1(zx)S_1(y)+2S_1(xyz)
\end{align}
Specializing to $z=-\frac{x+y}{2}$ yields a formula for $S_3(x+y)-S_3(x)-S_3(y)$ in terms of $S_1$.
A: I will use the standard notation $e_k$ for the $k$-th elementary symmetric polynomial of $n$ variables, and just $e$ instead of $e_1$ for better readability.
For $k\le n$, denote the set of the "base polynomials" of degrees $\leqslant k$ that are writable in terms of $x,y,e_1,e_2,\cdots,e_{k-1}$, possibly including coordinate-wise products as in the OP, by $P_{k-1}$.
For a partition $k=\ell+m$, define, in somewhat sloppy notation, the symmetric polynomial
$$E_{(\ell,m)}:=\sum (x_1\cdots x_\ell y_{\ell+1}\cdots y_k +y_1\cdots y_\ell x_{\ell+1}\cdots x_k ).$$
So the sum is over all pairs of disjoint $\ell$- and $m$-tuples of indices. (If $\ell=m$, each monomial occurs twice, but we will count it only once.)
For instance, the following sum has $8$ monomials, and it can be checked that we can decompose it into polynomials of $ P_{3}$ as follows:
$$E_{(3,1)}=\sum (x_1x_2x_3y_4+y_1y_2y_3x_4)=\\ e_3(x)e(y)+e_3(y)e(x)-[e_2(x)+e_2(y)]e(xy) \\ +e(x)e(xxy)+e(y)e(yyx)-e(xxxy)-e(yyyx).$$ (BTW this splits of course into two non-symmetric identities, one which contains all terms featuring three instances of $x$ and one of $y$, and the other one switching $x\leftrightarrow y$.)
Now, taking $k=4$, it is easy to see that we have $$e_k(x+y)-e_k(x)-e_k(y)=E_{(3,1)}+E_{(2,2)}$$ but it seems to me that 
$$E_{(2,2)}=\sum x_1x_2y_3y_4$$ can not be decomposed into polynomials of $ P_{3}$ . Or am I wrong? 
