I think the answer is no. Let $S={0}\cup[1,\infty)$ be the subsemiring of the (usual) reals. A non-zero ideal of $S$ is of the form ${0}\cup[a,\infty)$ where $a\geq 1$. Clearly, the only prime ideal of $S$ (according to your definition) is ${0}$ and it is subtractive. But no proper non-zero ideal of $S$ is subtractive.
Correction: My argument is not right: actually every non-zero ideal of $S$ ie either of the form ${0}\cup[a,\infty)$ or ${0}\cup(a,\infty)$ where $a\geq 1$. Hence $P={0}\cup(1,\infty)$ is a non-zero prime ideal which is not subtractive. And in general a unitary semiring has to always have a maximal ideal (as the unitary ring does.)
