First we notice that
\begin{split} 
a_n & = n \int_0^1 x^n (1+x)^{n-1}{\rm d}x \\
& = n \int_0^1 (1-x)^n (2-x)^{n-1}{\rm d}x \\
& = n\sum_{k=0}^{n-1} \binom{n-1}{k}2^k (-1)^{n-1-k} \int_0^1 (1-x)^n x^{n-1-k}{\rm d}x \\
&= \sum_{k=0}^{n-1} 2^k (-1)^{n-1-k} \binom{2n}{k} / \binom{2n}n. \\
\end{split}
Now, the numerator in the last expression is odd, and thus $\nu_2(a_n)=-\nu_2(\tbinom{2n}n)=-s(n)$ by [Kummer's theorem](https://en.wikipedia.org/wiki/Kummer%27s_theorem). QED