Are there Groups of Strictly Primes

As written your question seems to be underdetermined: you want a group law on the set $\mathcal{P}$ of prime numbers. As has already been pointed out, no problem: every nonempty set admits a group law, and for a countably infinite set this is especially easy to do (and does not require the Axiom of Choice): simply choose a bijection to your favorite countably infinite group and transport the group law using this bijection. Thus you can make a group with underlying set $\mathcal{P}$ which is: cyclic, finitely generated abelian but not cyclic, infinitely generated abelian, nilpotent of any desired nilpotency class, solvable but not nilpotent of any desired solvability class, simple,...If I am not mistaken the set of nonisomorphic group laws on $\mathcal{P}$ has the cardinality of the continuum.

However the first paragraph does not take into account that the elements of the countably infinite set are prime numbers in any way: any countably infinite set would work as well. You acknowledge that you want more than this, namely a natural group law on $\mathcal{P}$. But what do you mean by this? You don't say, so your question cannot really be answered in its current form.

There is a natural algebraic structure on $\mathcal{P}$, but it is not a group law. Rather, the natural structure is that $\mathcal{P}$ forms a basis for the commutative monoid $(\mathbb{Z}^+,\cdot)$ of positive integers under multiplication. This is precisely a restatement in more algebraic language of the uniqueness of prime factorization, and it has as a consequence that the monoid $(\mathbb{Z}^+,\cdot)$ is isomorphic to a free commutative monoid of countably infinite rank, namely $\bigoplus_{i=1}^{\infty} (\mathbb{N},+)$. (These considerations are worked out in more detail in the exercise G1) from the first problem set of an under/graduate number theory course I taught at UGA in 2007 and 2009: see here.)

In summary, the whole point of the prime numbers is that they are not closed under their natural operation. Your question is a bit like asking for a way of combining two chemical elements to get a third chemical element. Such operations can be defined, but are in a clear chemical sense unnatural: rather we combine elements to get compounds. That's the point of elements.

There are however natural group laws on quotient sets of the primes, especially if we generalize $\mathcal{P}$ to the set of prime ideals in a Dedekind ring $R$. Namely, let $\mathcal{I}(R)$ be the monoid of all nonzero ideals of $R$, and let $\operatorname{Cl} R$ be the ideal class group of $R$. Then we have natural maps

$\mathcal{P} \stackrel{\iota}{\rightarrow} \mathcal{I}(R) \stackrel{q}{\rightarrow} \operatorname{Cl} R$.

Now once again the map $\iota$ is an injection which realizes $\mathcal{P}$ as a basis for the free commutative monoid $\mathcal{I}(R)$: this is the natural algebraic structure on the primes! The map $q$ is a surjective monoid homomorphism. Denote by $\Phi$ the composite map $\mathfrak{p} \mapsto [\mathfrak{p}]$ which sends a prime ideal to its ideal class. Since $\mathcal{P}$ is a set of generators for $\mathcal{I}(R)$ and $q$ is surjective, $\Phi(\mathcal{P})$ is certainly a set of generators for $\operatorname{Cl} R$ (which is a group, but we get a set of generators even as a monoid -- i.e., without having to take inverses). One can ask whether or not $\Phi$ is actually surjective -- i.e., when every ideal class can be represented by a prime ideal. In this paper of mine I call $R$ replete when $\Phi$ is surjective. When this holds, the class group $\operatorname{Cl} R$ is a group formed out of the images of the prime numbers. Moreover, we can define something which is "almost" a group law on $\mathcal{P}$: we try to define $\mathfrak{p}_1 \cdot \mathfrak{p}_2 = \mathfrak{p_3}$ if $[\mathfrak{p_1}][\mathfrak{p_2}] = [\mathfrak{p_3}]$, but it doesn't quite work: $\mathfrak{p}_3$ is not well-defined but only its ideal class: i.e., if $\mathfrak{p}_4$ is another prime ideal such that there are $\alpha,\beta \in R \setminus \{0\}$ with $\alpha \mathfrak{p}_3 = \beta \mathfrak{p}_4$, then also $\mathfrak{p}_1 \cdot \mathfrak{p}_2 = \mathfrak{p}_4$. So it is a group law on the quotient of $\mathcal{P}$ by the equivalence relation of equality up to a principal ideal: it is still natural and important in the study of $R$ and $\operatorname{Cl} R$. When $R$ is not replete, the relationship between $\mathcal{P}$ and $\operatorname{Cl} R$ is more complicated.

When is $R$ replete? When $R = \mathbb{Z}_K$ is the ring of integers of a number field, then the repleteness of $R$ is a weak form of the Chebotarev Density Theorem (applied to the Hilbert class field of $K$), which says moreover that the fibers of the map $\Phi$ all have the same density. In my paper I observe that when $k$ is a field, $E_{/k}$ is an elliptic curve, and $k[E^{\circ}]$ is the coordinate ring of the affine curve $E^{\circ} = E \setminus \{O\}$, then $k[E^{\circ}]$ is (a Dedekind domain, as is well known) and is at least very close to being replete: it satisfies a condition that I called weakly replete which is good enough for the applications. But when is $k[E^{\circ}]$ actually replete? I give only partial information about this in Theorem 14 of my paper:

$\bullet$ If $k$ is alebraically closed, $k[E^{\circ}]$ is not replete.
$\bullet$ If $k$ does not have characteristic $2$ and $k[E^{\circ}]$ is not replete, then let $y^2 = P_3(x)$ be a Weierstrass equation for $E$. Then the $x$--coordinate map $E^{\circ}(k) \rightarrow k$ is surjective.

I don't say so in the paper, but the second part implies that $k[E^{\circ}]$ is replete whenever $k$ is a Hilbertian field of characteristic different from $2$: in particular when $k$ is a number field. And now that I look back on it, the characteristic not equal to $2$ hypothesis just looks like a bit of laziness on my part.

This discussion was a bit technical, but since you say you are interested in number fields and elliptic curves I thought you might be interested. In fact, giving a precise necessary and sufficient condition on $E$ and $k$ that makes $k[E^{\circ}]$ replete remains an open question. I am pretty sure that no one is working on this..

You can put a group structure on any nonempty set (this is equivalent to the axiom of choice). The question should be: does there exist a "natural" group structure on the set of all primes? In all humility, I don't believe that such a thing exists.

However, perhaps you'll be happy to hear this:

Let $\ell$ be a prime number, and let $\mathbf Z_\ell$ denote the ring of $\ell$-adic integers. It is a compact topological commutative ring. Let $\mathbb P$ denote the set of all prime numbers.

Theorem (Chebotarev - Dirichlet): the set $\mathbb P-\{\ell\}$ is dense in $\mathbf Z_\ell^\times$.

Corollary: Dirichlet's theorem on primes in arithmetic progressions.

I do believe, however, that the above theorem comes as close as possible to answering your question (this is my nice way of answering "no"). Primes give rise to conjugacy classes in Galois groups via the "Frobenius", so abelian Galois groups are a natural place to look if one would like to put a group structure on the set of all primes. The example above is the most natural one which appears in this context (because $\mathbf Z_\ell^\times$ appears as the Galois group of the infinite cyclotomic extension $\mathbf Q(\mu_{\ell^\infty})/\mathbf Q$).