Suppose $R$ is a ring. Then an element $x \in R$ is irreducible if it is not a unit and whenever $x=yz$ for some $y,z \in R$, then either $y$ or $z$ is a unit.
I'm reading a proof of a proposition which states that any non-zero element of a Noetherian domain is a product of irreducible elements. To prove this we start with an element $a$ and if it is not irreducible, then we factorize it as $a = yz$, with $y$ and $z$ not units. Why can we do that?
My thinking is (by negating the definition) that if $a$ is not irreducible then it either is a unit (can a unit always be factorized like that?), or when we can write it as $a=yz$ then $y$ and $z$ are not units.
Your objection is correct: there is an imprecision in the lemma you are stating. Some possible ways to phrase it correctly:
If $R$ is a Noetherian domain, then every non-zero, non-unit element of $R$ can be written as a finite product of irreducible elements.
If $R$ is a Noetherian domain, then every non-zero element of $R$ can be written as a product of a unit and finitely many irreducible elements.
If $R$ is a Notherian domain, then every non-zero element of $R$ is associated to a finite product of irreducible elements.
Where $(2)$ and $(3)$ are under the convention of $1$ being the empty product.
The author means one of those equivalent formulations, in which the case $a\in R^*$ is easily ruled out.