Yes, the ring of all algebraic integers has this property. Considered as a subring of the complex numbers, this is the set of all zeros of monic polynomials with integer coefficients. It is well known that this set forms a subring of the complex numbers.
Being a subring of the complex numbers, it is an integral domain. Thus, every prime element is irreducible. If an element of this ring is a product of primes, then it is a product of irreducibles. Thus, it suffices to show that the ring of all algebraic integers contains no irreducible elements.
Since the ring of all algebraic integers is not a field, we may choose a nonzero
nonunit algebraic integer $x$. Then $x = \sqrt{x} \cdot \sqrt{x}$ is a factorization into non-unit algebraic integers.
This ring does, however, have prime ideals. See here:
Prime ideals in the ring of algebraic integers
Edit: The ring of algebraic integers in the field $k = \mathbb{Q}(\sqrt{2}, \sqrt[4]{2}, \sqrt[8]{2}, \ldots)$ is an example where there exist irreducible elements (e.g., the rational integer 5) but in which not every element factors into a finite product of irreducibles (e.g., the rational integer 2 does not).
To see that $2$ does not factor as a finite product of irreducibles, note that if it did, then there would exist $n$ such that all of these irreducibles are contained in $\mathbb{Q}(\sqrt[n]{2})$. But in this field, the we have the ideal factorization $(2) = \left( \sqrt[n]{2} \right)^n$. It follows that an irreducible factor of $2$ could only be an associate of $\sqrt[n]{2}$, but these are not irreducible in $k$.
To see that $5$ is a prime element of $k$, note again that if the prime property failed, the elements involved would all exist in some common $\mathbb{Q}(\sqrt[n]{2})$, so $5$ would fail to be prime in this number field. It suffices, then, to show that the principal ideal $(5)$ is a prime ideal in the number field $F = \mathbb{Q}(\sqrt[n]{2})$.
Edit: Professor Lubin gives a short, elegant proof of this fact in his answer to this same post.
Alternatively, here is my original complicated argument. Consider the larger field $L = F(\zeta_{2^n})$ obtained by adjoining the $2^n$th roots of unity, which contains the subfield $K=\mathbb{Q} (\zeta_{2^n})$. Then $K/\mathbb{Q}$ is a cyclotomic extension, while $L/K$ is a Kummer extension. The following facts are well-known:
- The inertial degree of a rational prime in $K/\mathbb{Q}$ is equal to the order of the prime in the group of units of $\mathbb{Z}_{2^n}$.
- The number 5 generates a subgroup of this group of units of index $2$, so has order $2^{n-2}$ in the group.
In $K$, then, $5$ splits first as $(1+2i)(1-2i)$ in $\mathbb{Q}(i)$ and then remains inert the rest of the way up. If we show that $5$ continues to remain inert in $L/K$, then $5$ will have to be inert in $F/\mathbb{Q}$ since $F$ does not contain $\mathbb{Q}(i)$.
To see that the two prime ideals dividing $5$ are inert in the extension $L/K$, we note that $L/K$ is a tower of quadratic extensions with a totally ordered diagram of intermediate fields. It follows that if a prime is inert in the lowest quadratic extension $K(\sqrt[4]{2})$, then it remains inert throughout the extension $L/K$. Use some method to check that $5$ is inert in the extension $\mathbb{Q}(\sqrt[4]{2},\zeta_{16})/\mathbb{Q}(i)$. It follows that in the three quadratic extensions of $\mathbb{Q}(i)$ contained within $K(\sqrt[4]{2})$, the primes dividing $5$ do not split. The decomposition group of $5$ for the extension $K(\sqrt[4]{2})/\mathbb{Q}(i)$ must then be the entire Galois group, so the primes dividing $5$ are indeed inert in $K(\sqrt[4]{2})/K$ as claimed.