[I'm adding a new answer since what follows has almost nothing to do with my [previous one][1].]

Fix a real number $\alpha \ge 1$ and set $S_\alpha := \mathbb N \cup \mathbb R_{\ge \alpha}$. Of course, $S_\alpha$ is a commutative semidomain under the operations of addition and multiplication inherited from the real field, and the inclusion map $\mathbb N \to S_\alpha$ is a (unit-preserving) semiring embedding: The only unit of $S_\alpha$ is the (multiplicative) identity $1$; and it is easily checked that, for $\alpha > 1$, (the multiplicative monoid of the non-zero elements of) $S_\alpha$ is BF, by which we mean here that (i) every non-zero element of $S_\alpha$ is a product of atoms and (ii) the atomic factorizations of each element are bounded in length. In particular, it is seen that, if $\mathbb P \subseteq \mathbb N$ is the set of (natural) prime numbers, then the set $\mathscr A(S_\alpha)$ of atoms of $S$ is contained in the union of the (left-closed, right-open) interval $[\alpha, \alpha^2[$ with $[2, \alpha] \cap \mathbb P$: Most notably, we have that $\mathscr A(H) = \emptyset$ (and hence $S$ is not even atomic) when $\alpha = 1$, and $\mathscr A(S_\alpha) = [\alpha, \alpha^2[$ when $1 < \alpha \le 2$.

Now suppose that $1 < \alpha \le \sqrt{2}$. The set of atoms of $S_\alpha$ has then a "very smooth structure", and each prime numbers $p \in \mathbb N$ has a non-trivial factorization into atoms of $S_\alpha$ (since $\alpha \le \sqrt{2}$ and the only unit is the identity, $p$ is not an atom of $S_\alpha$); however, it is not clear to me if one can tune $\alpha$ so as to guarantee that such a factorization is essentially unique (possibly after replacing $\mathbb R_{\ge \alpha}$ with something smaller). In fact, I don't even know if $S_\alpha$ is FF, which means here that each non-zero element has only finitely many atomic factorizations that are pairwise non-equivalent in the obvious sense. 


  [1]: https://mathoverflow.net/a/415988/16537