Since this subject is full of misunderstandings (see here, here, here, and here) let us fix a precise terminology.
Let $A$ be a commutative ring and $P$ an $A$-module.

I) We'll say that $P$ is a locally free module of rank one or is invertible if $P$ is finitely generated, projective and of rank one in the sense that for every prime ideal $\mathfrak p$ of $A$ the localized $A_\mathfrak p$- module $P_\mathfrak p$ (which is free by projectiveness) is of dimension $1$.
These modules correspond bijectively, by a well known result of Serre in FAC, to locally free sheaves $\tilde P$ of rank $1$ on $\operatorname {Spec}A$, also known as invertible sheaves. This is one motivation for the above terminology.
Another justification for the terminology "invertible" is that these modules are exactly those for which the canonical evaluation map $ P^*\otimes_AP\to A$ is an isomorphism.

II) If $B\supset A$ is an overring of $A$ and $P\subset B$ is an $A$-module, we'll say that it is concretely invertible with respect to $B$ if $P.(A:P)_B=A$.
[As is standard $(A:P)_B$ denotes the set of elements $b\in B$ such that $bP\subset A$]
It is well-known that these concretely invertible modules are invertible. What about the converse?

Question:
Is an invertible $A$-module $P$ isomorphic as an $A$-module to a concretely invertible module $P'\subset B$ with respect to a suitable overring $B\supset A$?

Remarks
a) The answer is "yes" if $A$ is an integral domain. We can take $P'$ sitting inside $B=\operatorname {Frac}A$.
b) The answer is trivially yes if $A$ is semi-local, since then $P$ is free of rank $1$: see here.
c) According to Eisenbud's Commutative Algebra,Theorem 11.6 b., every invertible module $P$ over a noetherian ring $A$ is isomorphic to a concretely invertible submodule $P'\subset \operatorname {Quot} A$ of its total quotient ring $\operatorname {Quot} A$ obtained by inverting the regular (=not zero-divisors) of $A$
d) It is however not true in general that we can always find the required $P'$ inside the total quotient ring $B=\operatorname {Quot} A$.
Lam gives a counter-example in his Lectures on Modules and Rings, Example (2.22)(A), page 37.