Let $R$ be an integral domain. Let $R_0=R-\{0\}$ and $R^*$ be the unit group of $R$. An orientation of $R$ (my terminology) is a submonoid $N\subseteq R_0$ which intersects each associate equivalence class at exactly one point. That is, if $x\sim y$ (i.e. $x=uy$ for some $u\in R^*$) where $x,y\in N$ then $x=y$, and if $x\in R_0$ then there exists $y\in N$ such that $x\sim y$.
For example, $\Bbb Z^+$ is an orientation of $\Bbb Z$, and if $k$ is a field then the set of monic polynomials is an orientation of $k[X]$.
Does every integral domain have an orientation?
For PIDs, the answer is yes: For each prime ideal $P$, pick a prime generator $P=(p)$, and let $N$ be the set of products of these generators. Since $R$ is a UFD, no two distinct products can be associate, and moreover if $x\in R_0$ then $x$ is a product of primes, and for each prime $q_i$, if $p_i$ is the chosen generator of $(q_i)$ then $p_i\sim q_i$ so $x$ is associate to $\sum_ip_i$.
There are non-orientable commutative rings that are not integral domains. For example, in $\Bbb Z/6\Bbb Z$, there are only two units, so every square must be in $N$ (since either $x\in N$ or $-x\in N$ implies $x^2\in N$), so the only possible structure is $\{1,3,4\}$; but $3\cdot 4=0$ so this is not a submonoid.
If $R$ is not integrally closed there are problems. For example if $R=\mathbb{Z}[2i]$ then $R^\times=\{\pm1\}$, and $N$ will have to contain an associate of $2i$, and hence its square which is $-4$, and similarly an associate of $2$, and hence its square, which is $+4$.
If $R$ is integrally closed in its field of fractions $K$ then $R^\times$ is the torsion subgroup of $K^\times$, and the quotient $K^\times/R^\times$ is torsion-free. If it's also free then the sequence mentioned in my comment above splits, so this is a clue as to where to look for a counterexample in general; however my impression is that you're only interested in a yes/no thing and the first counterexample will do it.