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 $x\in A$.