The result is not true in general. Take for example an infinite dimensional, separable Hilbert space $\mathcal{H}$ and let $A=\mathcal{K(H)}$, the compact operators on $\mathcal{H}$. Note that $\mathcal{K(H)}\subset\mathcal{B(H)}$ is a non-degenerate representation. It is true that the pure states of $\mathcal{K(H)}$ are precisely the vector states $\omega_\xi$, where $\xi$ runs over the unit vectors on $\mathcal{H}$. A proof of this can be found in Murphy's book, example 5.1.1.
Now by the Krein-Milman theorem, we know that the weak-* closed convex hull of $\text{PS}(A)\cup\{0\}$ (where $\text{PS}(A)$ denotes the pure states on $A$) is equal to the set of positive linear functionals on $A$ of norm at most $1$. This is of course a proper subset of the dual space $A^*$.
But, if we go as far away as we can from the compacts, Glimm's lemma can give something relevant to what you guessed, (but not exactly $C=A^*$). Check out Brown-Ozawa, Chapter 1 for a statement and a very brief (and difficult imo!) proof of Glimm's lemma:
Under the hypothesis that $A\cap\mathcal{K(H)}=0$, each state on $A$ can be approximated in the weak-* topology by a net of vector states. Moreover, if $A$ is separable, then each state of $A$ can be approximated in the weak-* topology by a sequence of vector states $\omega_{\xi_n}$ so that $(\xi_n)$ is an orthonormal sequence. Now combining this with the fact that $A^*$ is the linear span of the states $S(A)$, gives something close to what you are describing here.
Edit
As (definitely rightfully) requested in the comments by QuantumSpace, I am showing that the weak-star closure of $\{\omega_\xi\}_\xi$ is equal to the state space, when the algebra is unital and non-degenerately represented. This is based on the answer of Ruy on this post. The "standard exercise" I am referring to can be proved easily by using a beautiful result of linear algebra: see this post.
Let $A\subset B(H)$ be a unital $C^*$-algebra that is non-degenerately represented, i.e. $1_A=\text{id}_H$. Then the set $F=\{\omega_\xi:\xi\in H, \|\xi\|=1\}$ is weak-star dense in the state space $S(A)$.
Of course $F\subset S(A)$ and let's take its weak-star closed convex hull, which I will call $E$, which is again a subset of $S(A)$, since the weak-star limit of states is a state, since $A$ is unital. Now assume that $\tau\in S(A)\setminus E$. By the Hahn-Banach separation theorem we can find a weak-star continuous functional $\phi:A^*\to\mathbb{C}$ and a real number $t$ so that
$$\text{Re}(\phi(\rho))<t<\text{Re}(\phi(\tau))$$
for all $\rho\in E$. It is a standard exercise that every weak-star continuous functional on $A^*$ is an evaluation functional, so $\phi=\text{ev}_a$ for some $a\in A$. By writing $a=x+iy$ for $x,y\in A_{sa}$ we have that $\phi(\sigma)=\text{ev}_a(\sigma)=\sigma(a)=\sigma(x)+i\sigma(y)$, so the real part of $\phi(\sigma)$ is equal to $\sigma(x)$ for all $\sigma\geq0$ in $A^*$. The inequality thus becomes
$$\rho(x)<t<\tau(x) $$
for all $\rho\in E$. In particular we have that
$$\langle x\xi,\xi\rangle_H<t=\langle t\cdot\text{id}_{H}\xi,\xi\rangle$$
for all unit vectors $\xi$, so, if $\eta\in H$ is arbitrary, then
$$\frac{1}{\|\eta\|^2}\langle x\eta,\eta\rangle<\frac{1}{\|\eta\|^2}\langle t\cdot\text{id}_H\eta,\eta\rangle$$
so we can wipe out the $1/\|\eta\|^2$ factor and thus $t\cdot\text{id}_{H}\geq x$ (as operators, i.e. this inequality takes place in $A$). But now $\tau$ is a state, so $\tau(x)\leq\tau(t\cdot\text{id}_H)=t\cdot\tau(1_A)=t$, a contradiction.
