Substitution for Landau's O notation formula
First let us notice that $\log s = n^{o(1)} = o(n)$, and so $\mathit{poly}(\log s, n) = \mathit{poly}(n)$. This shows that the first summand is $2^n \mathit{poly}(n)$. 

Next, since $\log s = n^{o(1)}$, also $O(\log s) = n^{o(1)}$ (since constants are $n^{o(1)}$) and $O(\log s)^c = n^{o(1)}$ (since a constant multiple of $o(1)$ is also $o(1)$), hence the second summand is $2^{n^{o(1)}} = o(2^n)$. So the first summand is more significant.