Here is how I think it should go. First

$\mathrm{Sym}\left(V\oplus \wedge^2 V\right)\cong \mathrm{Sym}\left(V\right)\otimes \mathrm{Sym}\left(\wedge^2 V\right)$

Then we have

$\mathrm{Sym}\left(\wedge^2 V\right)\cong \sum\limits_{\lambda} \mathrm{Schur}_{\lambda}\left(V\right)$

where the sum is over partitions such that all parts of the conjugate partition are even. This came up in 
http://mathoverflow.net/questions/13568/symmetric-tensor-products-of-irreducible-representations

The tensor product $\mathrm{Sym}\left(V\right)\otimes \mathrm{Schur}_{\lambda}\left(V\right)$ is known by Pieri's rule.

Now given a partition we take a maximal subdiagram such that every column has an even number of boxes. The complement is skew shape with at most one box in each column.