For the warm-up question, if $W$ is an infinite set, then $S_W=\{0\}$ whether $W$ contains a positive real or not, so let's assume it was intended that $W$ be finite.

If $\alpha$, not a positive real, is in $W$, then $p(x)=(x-\alpha)(x-\overline\alpha)$ vanishes at $\alpha$ and not at any positive real. More generally, if $W=\{\alpha_1,\dots,\alpha_m\}$ then $p(x)=\prod_j(x-\alpha_j)(x-\overline\alpha_j)$ vanishes on $W$ and not at any positive real.

I believe that if $p(x)$ has no positive real root then for $n$ sufficiently large $(x+1)^np(x)$ has no negative coefficients. This follows from a result cited in the accepted answer to Application of polynomials with non-negative coefficients. So the answer to the warm-up question (if I have interpreted that earlier answer correctly) is, yes, the converse is true.

For the warm-up question, if $W$ is an infinite set, then $S_W=\{0\}$ whether $W$ contains a positive real or not, so let's assume it was intended that $W$ be finite.

If $\alpha$, not a positive real, is in $W$, then $p(x)=(x-\alpha)(x-\overline\alpha)$ vanishes at $\alpha$ and not at any positive real. More generally, if $W=\{\alpha_1,\dots,\alpha_m\}$ then $p(x)=\prod_j(x-\alpha_j)(x-\overline\alpha_j)$ vanishes on $W$ and not at any positive real.

I believe that if $p(x)$ has no positive real root then for $n$ sufficiently large $(x+1)^np(x)$ has no negative coefficients. This follows from a result cited in the accepted answer to Application of polynomials with non-negative coefficients. So the answer to the warm-up question (if I have interpreted that earlier answer correctly) is, yes, the converse is true.