This is just an observation, not an answer. Too long for a comment.
See https://mathoverflow.net/questions/203033. The first (symmetric) part $S_k(x+y)$ can be generalised to any number of variable types x, y, z, .. by observing that the elementary summetric functions never produce a variable to a power >1. Hence we can 'encode' the variable types x, y, z, .. by $x^1, x^2, x^3, ..$ and obtain using 4 variables (=subscripts) of 3 types
$S_2(x+x^2+x^3)=x_2^3 x_1^3+x_3^3 x_1^3+x_4^3 x_1^3+x_2^2 x_1^3+x_3^2 x_1^3+x_4^2 x_1^3+x_2 x_1^3+x_3 x_1^3+x_4 x_1^3+x_2^3 x_1^2+x_3^3 x_1^2+x_4^3 x_1^2+x_2^2 x_1^2+x_3^2 x_1^2+x_4^2 x_1^2+x_2 x_1^2+x_3 x_1^2+x_4 x_1^2+x_2^3 x_1+x_3^3 x_1+x_4^3 x_1+x_2^2 x_1+x_3^2 x_1+x_4^2 x_1+x_2 x_1+x_3 x_1+x_4 x_1+x_2^3 x_3^3+x_2^2 x_3^3+x_2 x_3^3+x_2^3 x_4^3+x_3^3 x_4^3+x_2^2 x_4^3+x_3^2 x_4^3+x_2 x_4^3+x_3 x_4^3+x_2^3 x_3^2+x_2^2 x_3^2+x_2 x_3^2+x_2^3 x_4^2+x_3^3 x_4^2+x_2^2 x_4^2+x_3^2 x_4^2+x_2 x_4^2+x_3 x_4^2+x_2^3 x_3+x_2^2 x_3+x_2 x_3+x_2^3 x_4+x_3^3 x_4+x_2^2 x_4+x_3^2 x_4+x_2 x_4+x_3 x_4$
which is just this simple expression in monomial symmetric functions:
$m_{\{1,1\}}+m_{\{2,1\}}+m_{\{2,2\}}+m_{\{3,1\}}+m_{\{3,2\}}+m_{\{3,3\}}$
'Decoding' $x^1 -> x, x^2 -> y, x^3 -> z, ..$ recovers the original $S_2(x+y+z)$.
Generally, the following conjecture seems to hold: (using e for the elementary symm. pol.)
$e_k(v,t) = \sum m_\lambda$
where v gives the number of variables (=subscripts) and t gives the number of types used in the argument (x,y,z,..); the sum goes over all partitions $\lambda$ in k parts <= v, thus partitions of k up to k*v.