Formally, $S_4^-(a)$ is equal to $-a\,{}_5F_4(-a+1, -2a, -2a, -2a, -2a; -a, 1, 1, 1;1)$. This can be summed by Dougall's very well-poised $_5F_4$ summation formula (dlmf.nist.gov/16.4#E9) to give $$-\frac{a\Gamma(1+4a)}{\Gamma(1+2a)^3 \Gamma(1-2a)}.$$

These are sometimes called "vector partitions" or "bipartite partitions".

There is a product formula for the order polynomial of an unusual poset in "Solution of an Enumerative Problem Connected with Lattice Paths" by Kreweras and Niederhausen, sciencedirect.com/science/article/pii/S0195669881800200 though I don't know for sure that it's not included in those already mentioned. The poset is a product of a chain with a 3-element "V-shaped" poset.

@McRatchet To sum on $n$, use the binomial theorem. Specifically, if $n\ge a$ then $\sum_n \binom{n-a}{b}y^n = y^{a+b}/(1-y)^{b+1}$. For your second question, note that one of $j-k-1$ and $k-j$ is negative, so the expression is not a polynomial.

See also Richard Stanley's paper Invariants of finite groups and their applications to combinatorics, projecteuclid.org/euclid.bams/1183544328.

See Theorem 4 of my paper. (It solves $h_i(\vec{x}) = g_i(\vec{h})$; the variable $x$ is part of $g_i$.)