Something is wrong with the question, as here's a counter-example. Let $V=c_0$ with the pointwise involution (so this is a commutative C*-algebra). Let C be the obvious cone: the collection of vectors all of whose coordinates are positive. Let $x=(i,0,0,\cdots)$. Then $V^* = \ell^1$, so if $s=(s_n)\in\ell^1$ satisfies $s(C)\subseteq[0,\infty)$, we need that $s_n\geq 0$ for all $n$. But then $s(x)$ is purely imaginary!
So, maybe you also need $x^*=x$. Under this assumption, here's a proof, but it has nothing to do with "Krein-Milman"...
As C is closed, $V\setminus C$ is open, so let A be an open ball about x which doesn't intersect C. Then A and C are disjoint, non-empty, convex, so by Hahn-Banach, as A is open, we can find a bounded linear map $\phi:V\rightarrow\mathbb C$ and $t\in\mathbb R$ with $$ \Re \phi(a) < t \leq \Re \phi(c) $$ for $a\in A$ and $c\in C$. This is e.g. from Rudin's book. As $0\in C$, we see that $t\leq 0$.
Now, we can lift the involution * from V to the dual of V. In particular, define $$ \phi^*(x) = \overline{ \phi(x^*) } \qquad (x\in V)$$
So let $\psi = (\phi+\phi^*)/2$. For $c\in C$, as $c^*=c$, notice that $\psi(c) = \Re \phi(c)$. Hence $0 \leq \psi(c)$ for all $c\in C$. Similarly, as $x^*=x$, we have that $\psi(x) = \Re\phi(x)<t\leq 0$, as $x\in A$.