I based my answer on the following sources, check it out:
- [A-M] M. Atiyah; I. G. Macdonald, Introduction to Commutative Algebra.
- [H] T. Hungerford, Algebra (Springer, 1996).
- [S] J.-P. Serre, Corps Locaux.
Let $K$ be the field of fractions of your integral domain $R$. Your hypothesis is that for every nonzero prime ideal $P$ of $R$ we have $PP^{-1}=R$, where $P^{-1}$ is the set $(R:_KP)=\{z\in K: zP\subseteq R\}$. Since $R$ is an integral domain, then $R$ is contained in each one of its localizations, which in turn are contained in $K$.
Let $P$ be any nonzero prime ideal in $R$. Since $PP^{-1}=R$ then for some $p_1,\dots,p_m\in P$ and some $a_i\in P^{-1}$ we have $\sum_{i=1}^mp_ia_i=1$. Therefore for any $p\in P$ we have $p=p\cdot1=\sum_{i=1}^m(a_ip)p_i$, and since $a_ip\in R$ for each $i$, it follows that every prime ideal $P$ of $R$ is finitely generated, and it is well-known that this implies that $R$ is Noetherian (see for example [A-M], exercise 7.1).
Now consider $R_P$, the localization of $R$ at $P$ and let $Q=PR_P$ be its unique maximal ideal. Every element of $Q$ is of the form $p/t$, with $p\in P$ and $t\in R\setminus P$, and for any such element we have $a_i(p/t)=(a_ip)/t\in Q$, which shows that $a_i\in Q^{-1}$, and since $\sum_{i=1}^mp_ia_i=1$, then $1\in QQ^{-1}$. Thus, $Q$ is invertible as well. (This is Proposition 9.6 of [A-M], but the proof is flawed in the sense that previously the authors define quotients of submodules as ideals in the ring of scalars, whereas our inverse fractional ideals $I^{-1}$ are defined as subsets of the field of fractions. For this reason I chose to give a direct proof.)
We have $\cap_{n\geq1}Q^n=0$ by Krull's intersection theorem, and since $P\ne0$ then $Q\ne0$, so necessarily we have $Q\ne Q^2$. Let $a\in Q\setminus Q^2$. Then $aQ^{-1}\nsubseteq Q$ (otherwise $aQ^{-1}Q=aR\subseteq Q^2$), and since $aQ^{-1}$ is an ideal in $R_P$, which is a local ring with maximal ideal $Q$, it follows that $aQ^{-1}=R$, so $Q=aR$. I borrowed this argument from [H], Lemma VIII.6.9, fact v).
Consequently, $R_P$ is a Noetherian (because $R$ is Noetherian) local ring such that its maximal ideal $Q$ is generated by a non-nilpotent element (clear, because we are working on an integral domain). Now I invoke [S], Chapitre I, Proposition 2, which states that the conditions above characterize discrete valuation rings.
Thus, $R$ is a Noetherian integral domain such that $R_P$ is a discrete valuation ring for each nonzero prime ideal $P$ of $R$. It is well-known that this is equivalent to both of your desired conclusions. Such rings, as you probably know, are called Dedekind domains. For a proof of these and other interesting equivalences, see for example [H], Theorem VIII.6.10.
Best Answer
Yes. Factor $I = \displaystyle\prod_{i =1}^n \mathfrak{p}_i^{e^i}.$ Then by the Chinese Remainder Theorem $A/I \cong \displaystyle\bigoplus_{i =1}^n A/\mathfrak{p}_i^{e^i}.$ So it is enough to show each factor $A/\mathfrak{p}_i^{e^i}$ is principal. The ideals of $A/\mathfrak{p}_i^{e^i}$ are exactly the images of the ideals of $A$ containing $\mathfrak{p}_i^{e^i},$ i.e., $\mathfrak{p}_i^{n}$ for $1\le n \leq e_i,$ under the projection map $\pi:A \rightarrow A/\mathfrak{p}_i^{e^i}.$ If $\pi(\mathfrak{p}_i) = \pi(\mathfrak{p}_i^2)$ then $\pi( \mathfrak{p}_i) = 0$ and $A/\mathfrak{p}_i^{e_i}$ is a field. Otherwise, let $\alpha \in\pi( \mathfrak{p}_i) \setminus \pi(\mathfrak{p}_i^2).$ Then $(\alpha)$ is a proper ideal such that $(\alpha) \not\subset \pi(\mathfrak{p}_i^n) $ for any $n\ge 2.$ It follows $(\alpha) = \pi(\mathfrak{p}_i).$ We conclude $\pi(\mathfrak{p}_i^n) = (\alpha^n)$ and hence $A/\mathfrak{p}_i^{e_i}$ is principal.