Let $f$ be a symmetric function of $s$ variables. The identity is $$\sum_{all \ k's}^\infty f(k_1,k_2,k_3,...,k_s)=\sum_{n=s}^\infty \sum_{\lambda\vdash n}\frac{s!\prod_l \lambda_l}{z_\lambda} f(\lambda)$$ where $z_\lambda$ is the size of the centralizer of a permutation of type $\lambda$. So as you can see $\frac{s!\prod_l \lambda_l}{z_\lambda}$ is the number of compositions with the elements of $\lambda$. And don't forget that $\lambda$ is always a partition with $s$ parts.
I have verified it for some values of $s$. For example ($s$=3, expanding up to $n=6$):
$$ \sum_{k_1=1}^\infty\sum_{k_1=1}^\infty\sum_{k_1=1}^\infty f(k_1,k_2,k_3)=f(1,1,1)+3f(2,1,1)+[3f(2,2,1)+3f(3,1,1)]+[f(2,2,2)+6f(3,2,1)+3f(4,1,1)]... $$ So it is kind of obvious that the pattern emerges.
How can this be proved in a rigorous way? (for all $s\in \mathbb{N}$ off course!) Any reference on this kind of manipulations?
Also, if you have a better way to write the number $\frac{s!\prod_l \lambda_l}{z_\lambda}$ it will be helpfull.
[NOTE ABOUT FIRST VERSION]
In the first version of the question I wrote $$\sum_{all \ k's}^\infty x^{\sum_{j=1}^s k_j} f(k_1,k_2,k_3,...,k_s)=\sum_{n=s}^\infty x^n\sum_{\lambda\vdash n}\frac{s!\prod_l \lambda_l}{z_\lambda} f(\lambda)$$ But $x^{\sum_{j=1}^s k_j} f(k_1,k_2,k_3,...,k_s)$ is itself a symmetric function, so I just redefine $x^{\sum_{j=1}^s k_j} f(k_1,k_2,k_3,...,k_s)\to f(k_1,k_2,k_3,...,k_s)$
The $x$'s are not needed and now the identity has a simpler form.
