Though two answers are already posted, let me explain how to understand that it is false from general reasoning. The inequality $\|(I+A+B)^{-1}A\|\leqslant 1$ is equivalent to $\|(I+A+B)^{-1}A x\|\leqslant \|x\|$ for any vector $x$. Take $x=A^{-1}(I+A+B)y$, the inequality rewrites as $\|y\|\leqslant \|y+A^{-1}(I+B)y\|$. We are given only that $A,B$ are positive definite, thus for given vector $z$, $A^{-1}z$ may be any vector which forms an acute angle with $z$. For $z=(I+B)y$ this in general does allows the vector to form an obtuse angle with $y$ (unless $y$ is an eigenvector of $B$.) So, if direction of $A^{-1}(I+B)y$ forms an obtuse angle with $y$ and $A^{-1}$ is small enough, the inequality $\|y\|\leqslant \|y+A^{-1}(I+B)y\|$ fails.

Though two answers are already posted, let me explain how to understand that it is false from general reasoning. The inequality $\|(I+A+B)^{-1}A\|\leqslant 1$ is equivalent to $\|(I+A+B)^{-1}A x\|\leqslant \|x\|$ for any vector $x$. Take $x=A^{-1}(I+A+B)y$, the inequality rewrites as $\|y\|\leqslant \|y+A^{-1}(I+B)y\|$. We are given only that $A,B$ are positive definite, thus for given vector $z$, $A^{-1}z$ may be any vector which forms an acute angle with $z$. For $z=(I+B)y$ this in general allows the vector to form an obtuse angle with $y$ (unless $y$ is an eigenvector of $B$.) So, if the direction of $A^{-1}(I+B)y$ forms an obtuse angle with $y$ and $A^{-1}$ is small enough, the inequality $\|y\|\leqslant \|y+A^{-1}(I+B)y\|$ fails.