Is every locally free module of rank $1$ over a commutative ring concretely invertible?

The answer is yes. Recall that given an invertible $A$-module $P$ and $n \in \mathbf{Z}$ there is an invertible $A$-module $P^{\otimes n}$ such that $P^{\otimes 0} = A$, $P^{\otimes 1} = P$, and $P^{\otimes n} \otimes_A P^{\otimes m} = P^{\otimes n + m}$. Set $B = \bigoplus_{n \in \mathbf{Z}} P^{\otimes n}$; this is a commutative $\mathbf{Z}$-graded $A$-algebra (details omitted). Then $P \subset B$ in degree $1$ and $(P : A)_B = P^{\otimes -1}$ sitting in degree $-1$ and we have $P \cdot P^{\otimes -1} = A$.

Remark. The spectrum of $B$ is the $\mathbf{G}_m$-torsor over $\text{Spec}(A)$ corresponding to $P$.

This is a partial answer of a more general problem.

If $A$ is a commutative ring with few zerodivisors (which holds iff $\text{Quot}(A)$ is semilocal), then the answer is yes: in that case, an invertible $A$-module is equivalently a concretely invertible $A$-submodule of $\text{Quot}(A)$. This result includes the cases you mention where $A$ is a domain or $A$ is semilocal or $A$ is Noetherian. For an outline of the proof, see Exercises 14 and 15 of Section 2.5 of my new book, "Rings, Modules, and Closure Operations." (Sorry, this isn't meant as a shameless plug :) I suspect that in general one needs to use the complete ring of quotients $Q(R)$. In other words, I'd conjecture that, for a general commutative ring $A$, an $A$-module is invertible iff it is a concretely invertible $A$-submodule of $Q(A)$. (This might even generalize to noncommutative rings.)

EDIT (to clear up some confusion):

In the answer above, $Q(R)$ denotes the complete ring of quotients, also known as the maximal ring of quotients, which in general contains but can be much bigger than $\text{Quot}(R)$. See planetmath.org/completeringofquotients. Unlike the total quotient ring, the maximal ring of quotients is defined even for noncommutative rings (although then one must distinguish between the maximal left ring of quotients and the maximal right ring of quotients).

ANOTHER EDIT: I thought the original question was to find an overring $B$ of $A$ that every invertible $A$-module is isomorphic to some concretely invertible $A$-submodule of $B$. This is what the total quotient ring $B = \text{Quot}(A)$ achieves in the case where $A$ is a ring with few divisors, and I suspect that this is what the complete ring of quotients $B = Q(A)$ achieves for any commutative ring $A$. The answer that was accepted, however, doesn't require that a single overring $B$ work for every $P$. The problem of determining a singe $B$ that works for every $P$ is still open.