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, for general background on meromorphic functions on arbitrary schemes see EGA IV$_4$, sec. 20, esp. 20.1.3, 20.1.4. (There is an error: in (20.1.3), $\Gamma(U,\mathcal{S})$ should consist of locally regular sections of $O_X$; this is the issue in the Kleiman reference mentioned by Georges. I promise that the content of EGA works just fine upon making that little correction; I checked it a long time ago. There are more hilarious errors in there, all correctable, such as fractions with infinite numerator and denominator, but that's a story for another day). Also, 20.2.12 is the result cited from Liu's book.
In the setup in the question, it should really say "we could have invertible meromorphic functions on Spec($A$) that don't come Frac($A)^{\times}$". This is what I will now prove cannot happen.
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