Hopefully there are no errors in the proof below that no such exotic invertible meromorphic functions exist in the affine case; it will be nice if someone can confirm the correctness. The argument below is a correction on an earlier attempt which had a bone-headed error. [Kleiman's construction from Georges' answer is not invertible, so no inconsistency. Kleiman makes some unfortunate typos -- his $\oplus k(Q)$ should be $\prod k(Q)$, and more seriously the $t$ at the end of his construction should be $\tau$, for example -- but not a big nuisance.]
First, for anyone curious about general background on meromorphic functions on arbitrary schemes, see EGA IV$_4$, sec. 20, esp. 20.1.3, 20.1.4. (There is a little subtle 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. The content of EGA works just fine upon making that little correction. There are more hilarious errors elsewhere in IV$_4$, 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 $U_i$; we may and do assume each $U_i$ is quasi-compact. 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 $J=A$.
Since the zero scheme $V({\rm{Ann}}(s_i))$ is disjoint from $U_i$ (as $s_i|_ {U_i}$ is a regular section), the closed sets $V({\rm{Ann}}(s_i))$ and $V({\rm{Ann}}(s_2))$ have intersection disjoint from $U_1 \cup U_2$. In other words, the quasi-coherent ideals ${\rm{Ann}}(s_1)$ and ${\rm{Ann}}(s_2)$ generate the unit ideal over $U_1 \cup U_2$. A quasi-coherent sheaf is generated by global sections over any quasi-affine scheme, such as $U_1 \cup U_2$ (a quasi-compact open in an affine scheme), so we get $a_1 \in {\rm{Ann}}(s_1)$ and $a_2 \in {\rm{Ann}}(s_2)$ such that $a_1 + a_2 = 1$ on $U_1 \cup U_2$. Multiplying both sides by $s_1 s_2$, we get that $s_1 s_2 = 0$ on $U_1 \cup U_2$. But $s_1$ is a regular section over $U_1$, so $s_2|_ {U_1} = 0$. But $s_2|_ {U_2}$ is a regular section, so we conclude that $U_1$ and $U_2$ are disjoint. This argument shows that the $U_i$ are pairwise disjoint.
Thus, {$U_i$} is a finite disjoint open cover of Spec($A$), so in fact each $U_i = {\rm{Spec}}(A_i)$ with $A = \prod A_i$. But recall that in $A$ every non-unit is a zero-divisor. It follows that the same holds for each $A_i$ (by inserting 1's in the other factor rings), so each regular section $s_i|_ {U_i} \in A_i$ is a unit. But the preceding argument likewise shows that $s_i|_ {U_j} = 0$ in $A_j$ for $j \ne i$, so each $s_i \in A$ has a unit component along the $i$th factor and vanishing component along the other factors. Hence, the $s_i$ generate 1, so $J = A$. QED