Hmm, I surrender and post my long string of comments as an answer (and have now deleted the long string of comment boxes). Hopefully there are no bone-headed errors in the affirmative proof below that no such exotic invertible meromorphic functions exist in the affine case; it will be nice if someone can confirm the correctness (e.g., the OP?).
First, see EGA IV$_4$, sec. 20, esp. 20.1.3, 20.1.4 (error: in (20.1.3), $\Gamma(U,\mathcal{S})$ should consist of locally regular sections of $O_X$). Also, 20.2.12 is the result from Liu's book. The setup in the question should really say "we could have invertible meromorphic functions on Spec($A$) that don't come Frac($A)^{\times}$". The first step is the observation that for any scheme $X$, the ring $M(X)$ of meromorphic functions is naturally identified with the direct limit of the modules Hom($J, O_X)$ as $J$ varies through ideals which contain a regular section Zariski-locally on $X$. Basically, such $J$ are precisely the quasi-coherent "ideals of denominators" of global meromorphic functions. This description of $M(X)$ is left to the reader as an exercise, or see section 2 of the paper "Moishezon spaces in rigid-analytic geometry" on my webpage for the solution, given there in the rigid-analytic case but by methods which are perfectly general.
Now working on Spec($A$), a global meromorphic function "is" an $A$-linear map $f:J \rightarrow A$ for an ideal $J$ that contains a non-zero-divisor Zariski-locally on $A$.
Assume $f$ is an invertible meromorphic function: there are $s_i \in J$ and a finite open cover $\{U_i\}$ of Spec($A$) (yes, same index set) so that $s_i$ and $f(s_i)$ are non-zero-divisors on $U_i$. Let $S$ be the non-zero-divisors in $A$. Hypotheses are preserved by $S$-localizing, and it suffices to solve after such localization (exercise). So without loss of generality each element of $A$ is either a zero-divisor or a unit, and hence all non-units of $A$ lie in minimal primes (see Bourbaki, Comm. Alg., Ch. IV, Exer. 17(b)). If $J=A$ then $f(x)=ax$ for some $a \in A$, so $a s_i=f(s_i)$ on each $U_i$, so all $a|_{U_i}$ are regular, so $a$ is not a zero-divisor in $A$, so $a$ is a unit in $A$ (due to the special properties we have arranged for $A$). Hence, it suffices to show $J=A$.
Now observe that we can assume $J$ is finitely generated: we may replace it with the ideal generated by the $s_i$. Without loss of generality, each $s_i$ is a non-unit, hence $s_i$ lies in some minimal prime $P_i$ for each $i$. Then $J$ is contained in the union of these $P_i$, so by finitude of this set of primes it follows that $J$ is contained in some $P_{i_0}$ (Prop. 1.11(i), Atiyah-MacDonald). That is, $J$ is contained in a minimal prime $P$. But $\{U_i\}$ covers Spec($A$), so some $U_ {i_1}$ contains the point $P$ in Spec($A$). Then $s_ {i_1}$ is regular at $P$ (as it's regular over $U_ {i_1}$), yet $s_ {i_1} \in P$. By minimality of $P$, all elements of the maximal ideal of $A_P$ are nilpotent, but $s_ {i_1}$ is such an element and yet is also not a zero divisor in this local ring. Contradiction!
Thus, we conclude that indeed $J = A$, so we win. QED