The claim is true. Lets start by simplifying the expression as follows:
$$x\left(1+\frac{1}{1+s_1x+s_2x^2+s_3x^3+\cdots}\right)=\frac{\sum_{n\geq 1}s_{2n-1}x^{2n}}{\sum_{n\geq 0} s_{2n} x^{2n}}.$$
This follows because $\sum_{n\geq 0}s_{2n}x^{2n}=\left(\sum_{n\geq 0}s_nx^n\right)^2$, and some basic manipulations over $\mathbb F_2$.

So we are interested in the coefficient of $x^{2n-1}$ in 
$$1+\frac{1}{1+s_1x+s_2x^2+\cdots}=\left(\sum_{n\geq 0}s_nx^n\right)+\left(\sum_{n\geq 0}s_nx^n\right)^2+\left(\sum_{n\geq 0}s_nx^n\right)^3+\cdots,$$
but by taking binary expansions we see that each term is of the form
$$\left(\sum_{n\geq 0}s_{2^{i_1}n}x^{2^{i_1}n}\right)\cdots \left(\sum_{n\geq 0}s_{2^{i_k}n}x^{2^{i_k}n}\right),$$
where $i_1,i_2,\dots,i_k\geq 1$. It follows that the terms in the coefficient of $x^{m}$ correspond exactly to the solutions of $m=j_1+2j_2+\cdots 2^rj_{r+1}$. The ones that appear with a nonzero coefficient in $\mathbb F_2$ are precisely the ones where the $j_1,j_2,\dots$ are all odd. So the number of terms in $p_n$, with your notation, is the number of ways we can write 
$$2n-1=\sum_{r= 0}^k (2i_r+1)2^{r}$$
 for some k. This is the same as $$n=2^k+\sum_{r= 0}^k i_r 2^r$$ 
which gives exactly the number of ways to partition $n$ as powers of $2$. For example when $n=6$, we can write $11$ as $1+10, 1+6+4, 5+6, 5+2+4, 9+2,11$ each corresponding to the terms in $p_6=s_1s_5^2+s_1s_3^2s_1^4+s_5s_3^2+s_5s_1^2s_1^4+s_9s_1^2+s_{11}$.