I am trying to establish some implications between being an ID, a PID, a Noetherian ring and a valuation ring.
First of all, I know that PID $\Rightarrow$ Noetherian, because in a PID every ideal is generated by only one element, thus every ideal is finitely generated so the ring is Noetherian.
Now, I think that to go the other way we must ask for the ring to also be a valuation ring, i.e., I think that for a ring $A$ $$\text{Noetherian} + \text{Valuation ring}\Longleftrightarrow \text{PID}$$ My proof for $\Rightarrow $ is this. Let $I\subset A$ be an ideal of $A$. Then $I=(a_1,\ldots ,a_n)$. To make it simpler, I assume $n=2$. Since the ring is a valuation ring, either $a_1/a_2\in A$ or $a_2/a_1\in A$. Thus, we can get rid of one of the generators of $I$ and therefore conclude that $A$ is a PID.
So then I came up witht the following question. Every PID is an ID. So what conditions do we need to add to being Noetherian in order to get an ID? Of course, if we ask that our Noetherian ring is also a valuation ring, then it is an ID (in particular because every valuation ring is an ID). But I am looking for the weakest condition possible.
No, not necessarily. There are clearly PIDs that aren't valuation domains, like $F[x]$.
I think the condition you are fishing for is "Bezout domain." A commutative ring is called Bezout if each finitely generated ideal is principal. One can say: A Noetherian domain is a PID iff it is a Bezout domain." Valuation domains are all Bezout domains.
The conditions of being Noetherian and being a domain are quite "orthogonal" in the sense that there's no logical implication between the two. I can't imagine you'll find anything weaker than "domain" to add to Noetherian to get an integral domain.
Usually one only asks a question like "what do we need to add to P to get Q?" is if Q implies P and one is looking for the reverse relationship between the two.
There is a chart here on my website summarizing some relationships between domains.