[I'm adding a new (non-)answer since what follows has almost nothing to do with my previous one.]
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. Further, 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 $S_\alpha$ 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 $S_\alpha$ 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" (so I would consider it to be "simple" in the vague sense of condition (3) of the OP), and each $p \in \mathbb P$ 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,
Note that, in general, $S_\alpha$ is not FF, which means here that each non-zero element has only finitely many atomic factorizations that are pairwise non-equivalent in the obvious sense: If, for instance, $\alpha = \sqrt{2}$, then $a$ and $3a^{-1}$ are atoms of $S_\alpha$ (and hence the $S_\alpha$-word $a \ast 3a^{-1}$ is a length-$2$ atomic factorization of $3$ in $S_\alpha$) for all $a \in \bigl]\frac{3}{2}, 2\bigr[$ (see my answer here if something in the terminology or in the notation doesn't sound familiar).