This question is related to and inspired by a previous question What is the residue obtained from this integral? , but note that the appearing functions are slightly different.
Consider the following function in two complex variables $z_1$ and $z_2$:
$$F(z_1,z_2)=\frac{1}{z_1 z_2}\frac{1}{ f_1(z_1,z_2)}\frac{1}{ f_2(z_1,z_2)}$$
where functions $f_1$ and $f_2$ are defined as:
$$f_1(z_1,z_2)=a+z_1+\frac{z_1}{z_1+z_2}~~~~~,~~~~~f_2(z_1,z_2)=b+z_2+\frac{z_2}{z_1+z_2}$$
The global residue theorem states that the sum of all residues gives zero for any function that is holomorphic everywhere up to a finite amount of points (since, by contour deformation, this corresponds to an integral over a closed contour with no poles enclosed). Let us denote by $R_F([z_1],[z_2])$ the residue obtained from function $F(z_1,z_2)$ while localizing to zero the poles $[z_1]$ and $[z_2]$ from the denominator, which are associated with varying $z_1$ and $z_2$ respectively. With this, the global residue theorem yields for $F(z_1,z_2)$:
$$R_F(f_1,f_2)+R_F(z_1,f_2)+R_F(f_1,z_2)+R_F(z_1,z_2)=0$$
In this case there are no poles at infinity. Now it is relatively easy to check that we get:
$$z_1=\frac{-a(1+a+b)}{a+b}~~~,~~~z_2=\frac{-b(1+a+b)}{a+b}~~~\Rightarrow ~~~R_F(f_1,f_2)=\frac{a+b}{ab(1+a+b)}$$ $$z_1=0~~~,~~~z_2=-1-b~~~\Rightarrow ~~~R_F(z_1,f_2)=\frac{-1}{a(1+b)}$$ $$z_1=-1-a~~~,~~~z_2=0~~~\Rightarrow ~~~R_F(f_1,z_2)=\frac{-1}{b(1+a)}$$
The final residue at $z_1=0$ and $z_2=0$ depends on what $a$ and $b$ are. But it can be only one of the two following values:
$$R_F(z_1,z_2)=\frac{1}{a(1+b)}~~~~~~~or~~~~~~~R_F(z_1,z_2)=\frac{1}{b(1+a)}$$
Clearly, the sum of the four terms does not vanish. It is interesting to note that the two possibilities in $R_F(z_1,z_2)$ cancel either $R_F(f_1,z_2)$ or $R_F(z_1,f_2)$, but the first residue $R_F(f_1,f_2)$ is never cancelled by anything. What went wrong?