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The first step in the proof 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. 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 $J=A$.

The first step in the proof 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 quasi-coherent ideals which contain a regular section of $O_X$ 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 finitely many $s_i \in J$ 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. 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 $J=A$.

In the setup in the question, it should really say "we could have invertible meromorphic functions on Spec($A$) that don't come Frac($A)^{\times}$", since those are what give rise to "extra divisors". This is what I will prove cannot happen. The argument 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.]

In the setup in the question, it should really say "we could have invertible meromorphic functions on Spec($A$) that don't come Frac($A)^{\times}$", since those are what give rise to "extra principal Cartier divisors". This is what I will prove cannot happen. The argument 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.]

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 there is the result cited from Qing Liu's book in the setup for the question.

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.

In the setup in the question, it should really say "we could have invertible meromorphic functions on Spec($A$) that don't come Frac($A)^{\times}$", since those are what give rise to "extra divisors". This is what I will prove cannot happen. The argument 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.]

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 there is the result cited from Qing Liu's book in the setup for the question.

The first step in the proof 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.