Edit: The answer below is incorrect. While it is true that, via class field theory, we can recover the class group as a quotient of $G^{ab}$, the problem, as @ThePiper points out, is that this quotient is by $\widehat{\mathcal O}_K^\times$, which $G^{ab}$ knows nothing about.
Given the whole of $G$, we would be able to recover $\widehat{\mathcal O}_K^\times=\prod_{v}\widehat{\mathcal O}_{K_v}^\times$ via class field theory if we could recover the inertia groups $I_v$ from $G$: by local class field theory, $I_v\cong {\mathcal O}_{K_v}^\times$.
It is possible to recover the inertia groups from $G$. However, the fact that we can do so is a key part of the Neukirch-Uchida theorem.
The answer is yes. Let $G^{ab}$ denote the abelianisation of $G$ $-$ i.e. $G^{ab} = G/\overline{[G,G]}$. By global class field theory, we have a canonical isomorphism
$$K^\times\backslash\mathbb A_K^{\times}/\overline{(K_\infty^\times)^0}\cong G^{ab}.$$
Here, $\mathbb A_K^\times$ are the ideles of $K$, and $\overline{(K_\infty^\times)^0}$ is the closure of the identity connected component of $(K\otimes_\mathbb Q\mathbb R)^\times$ viewed as a subgroup of $\mathbb A_K^\times$.
This isomorphism gives a concrete connection to the class group of $K$: the class group of $K$ is canonically isomorphic to
$$K^\times\backslash\mathbb A_K^{\times}/\widehat{\mathcal O_K^\times} K_\infty^\times,$$
and is therefore a quotient of $G^{ab}$.
On the Galois side, this quotient of $G^{ab}$ cuts out a finite abelian extension of $K$ -- the Hilbert class field.
Assuming it is question of a quadratic extension of number fields, $K/\Bbb Q$ is Galois, and a prime $p\in \Bbb Z=\mathcal O_{\Bbb Q}$ splits into $(P_1 P_2 \dots P_r )^e$ in $\mathcal O_K$ where the $P_i$ are distinct primes, all having the same inertial degree $f$
over $p$. Moreover $ref = [K : \Bbb Q ]=2$.
You listed all the possible splitting behaviours of $p\mathcal O_K$.
Now taking a random prime ideal $P\in \mathcal O_K$, it is not always of the form $p\mathcal O_K$ for a prime $p\in \Bbb Z$, as an example:
Take $K=\Bbb Q(i)$, and $P=(1-2i)$ is a prime of $\mathcal O_K$ lying over $5=(1-2i)(1+2i)$. But $P$ is not of the form $p\mathcal O_K$ for $p\in \Bbb Z$ prime.
Best Answer
Yes, every ideal class contains a split prime. Here is one possible proof, which however doesn't satisfy the conditions of part B of your question: the analogue of Dirichlet's argument, using ideal class characters, shows that $\sum_{\mathfrak p \in [I]} N\mathfrak p^{-1}$ (the sum being taken over prime ideals in the class of $I$) diverges. On the other hand, the sum over non-split primes converges (because a non-split prime lying over $p$ has norm at least $p^2$, and so the sum over non-split primes is majorized by (some constant times) $\sum_p p^{-2}$, which converges). Thus there must be infinitely many split primes contributing to this sum.
Note: As the OP points out in a comment below, this argument applies only in the Galois case, and so doesn't actually address the question.