This is just a partial answer.
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 $\text{Quot}(R)$. This is defined in one of Lam's books somewhere, or 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).