Let $D$ be a Noetherian integral domain which is not a UFD. Being Noetherian guarantees that every non-zero non-invertible element $d \in D$ can be written as a product of irreducible elements of $D$, see this question; of course, such a product is not necessarily unique (for example, $(1+\sqrt{3}i)(1-\sqrt{3}i)=4=2 \cdot 2$).
It is not difficult to show that if some element $d \in D$ can be written as a product of prime elements of $D$, then this product is unique (if $d=p_1 \cdots p_m= u_1 \cdots u_n$, where the $p_i$'s are primes and the $u_i$'s are irreducibles, then use the primality of each $p_j$ and the fact that $D$ is an integral domain).
I wish to find a Noetherian integral domain $D$ and an element $d \in D$ such that $d= v_1 \cdots v_l$ is unique, $v_1 \ldots, v_l$ are irreducibles, with at least one of the $v_j$'s non-prime. (Hence the converse of the above fact is not true). Am I missing an easy example? (probably). What if we remove the assumption that $D$ is Noetherian?
(Remark: If some of the $v_j$'s are primes, then they must appear in every decomposition of $d$, but here this fact is not relevant, since I have already assumed that $d$ has a unique decomposition).
I think you can use the example you consider in your post, namely $D = \mathbb{Z}[i \sqrt{3}]$. Consider the factorization $$w = 2 \cdot (1 + i\sqrt{3}).$$ The generic element of our domain is $z=a + i \sqrt{3} b$ where $a,b \in \mathbb{Z}$ and this has absolute value squared $|z|^2 = a^2 + 3b^2$. Only certain positive integers take this form, and we can, with a bit of work, write down all the $z$ with a given absolute value squared.
Suppose we factor $w$ into irreducibles $w = z_1 \cdots z_n$. Then we must have $|z_1|^2 \cdots |z_n|^2 = 16$. This constraint leaves only finitely many possiblities for $z_1,\ldots,z_n$ (Note in particular that any $z_i$ with $|z_i|^2 = 1$ is one of the two units $\pm 1$). Some casework should allow you conclude that the above factorization of $w$ into irreducibles is unique, up to units.
However, neither $2$ nor $1+i\sqrt{3}$ is prime, as can be seen from your post.