Consider the following problem
Let $R$ be an integral domain and $R^{\times}$its group of units. An element of $S=R \backslash\left(R^{\times} \cup\{0\}\right)$ is irreducible if it is not a product of two elements in $S$. When $R$ is Noetherian, show that every element of $S$ is a product of finitely many irreducible elements of $S$.
The way that I thought about this problem was in the following way:
$S$ is the quotient of a group with a normal subgroup so it is a group. As $R$ is Noetherian, then extending $S$ as a group should also be Noetherian. Now assume that there is an element that is an infinite product of irreducibles. We need contractadict the assending chains condition. This is the same as contradicting the descending chains condition. Say $$ u= a_1 ... a_n ... . + R^{\times} \cup \{ 0\} $$ Now define $$ I_n = (a_1 ... a_n + R^{\times} \cup \{ 0\}). $$ Clearly this is a descending chain. So there exists $N$ such that for all $n>N$ we have $I_n= I_N$. This gives us that given $n$ there exists $d$ such that $$ a_1 a_2 ... a_n + R^{\times} \cup \{ 0\} = a_1 ... a_N \cdot d + R^{\times} \cup \{ 0\} $$ meaning $$ -a_1 a_2 ... a_n +a_1 ... a_N \cdot d\in R^{\times} \cup \{ 0\}. $$ From here on I got stuck as I wasn't able to get a contradiction.
I am aware that this methodology is not even correct as I do not think that $R^{\times} \cup \{ 0\}$ is an ideal because if it was then it will be whole set. So clearly I can't talk about $S$ outside of the sense that is a group.
Also, I do not understand what "is not a product of two elements of $S$" means as $S$ is a group under addition. Is this to do with the inherited $\cdot$ from the ring?
If anybody could help me with this I would greatly appreciate it.