Bezout’s identity for analytic functions of several variables

Make an open cover $D^2=\cup_j(U_j\cup V_j)$, for example, by polydisks such that $f$ has zeros only in $U_j$ and $g$ has no zeros in $U_j$. This is possible since zeros of $f$ and $g$ are disjoint.

Solve the 1st Cousin problem with Cousin data $-1/(fg)$ in $U_j$ and $0$ in $V_j$. The solution is a meromorphic function $\phi$ such that $\phi+1/(fg)$ is holomorphic in $U_j$ and $\phi$ is holomorphic in $V_j$. Let $v:=-f\phi$. Then $-v+1/g$ is holomorphic and divisible by $f$ in $U_j$, and thus $v$ is also holomorphic in $U_j$ since $1/g$ is holomorphic in $U_j$. So $v$ is holomorphic everywhere. Now $-v+1/g$ is divisible by $f$ also in $V_j$ since $f$ has no zeros in $V_j$. Then since $-vg+1$ is holomorphic and divisible by $f$, then $u:=(1-vg)/f$ is holomorphic and $uf+vg=1$ as required.

In modern texts they refer to H. Cartan's theorems A and B, but the case of polydisk of dimension 2 this was in the original paper of Cousin.

For the record, let me translate Alexandre Eremenko's answer in modern terms (after all, this is why sheaf theory was invented). The hypothesis implies an exact sequence $$0\rightarrow \mathscr{O}_{\mathbb{D}}\xrightarrow{\ (-v,u)\ } \mathscr{O}_{\mathbb{D}}^2\xrightarrow{\ (u,v)\ } \mathscr{O}_{\mathbb{D}}\rightarrow 0$$hence, using $H^1(\mathbb{D}^2,\mathscr{O})=0$ (Cartan's theorem B), the map $(u,v):H^0(\mathbb{D}^2,\mathscr{O})^2\rightarrow H^0(\mathbb{D}^2,\mathscr{O})$ is surjective.