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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.