The claim is false. Consider the following matrix argument.

\begin{eqnarray*} \|(I+A+B)^{-1}A\| \le 1\quad\Leftrightarrow\quad \begin{bmatrix}I & (I+A+B)^{-1}A \\ (I+A+B)^{-1}A & I \end{bmatrix} \ge 0. \end{eqnarray*} The latter inequality amounts to showing $I \ge (I+A+B)^{-1}A^2(I+A+B)^{-1}$, or in other words, we must show that $(I+A+B)^2 \ge A^2$. Since the map $x \mapsto x^2$ is not operator monotone, one can likely make this inequality fail.

Here is an explicit counterexample: \begin{equation*} A = \begin{bmatrix}1 &5\\ 5 &61 \end{bmatrix},\quad B = \begin{bmatrix}9 &-9\\ -9 & 13\end{bmatrix},\quad C = (I+A+B)^2-A^2 = \begin{bmatrix}111 &-654\\ -654& 1895 \end{bmatrix}. \end{equation*} The matrix $C$ has a negative eigenvalue: $1003 - 2\sqrt{305845} < 0$.

The claim is false. Consider the following matrix argument.

\begin{eqnarray*} \|(I+A+B)^{-1}A\| \le 1\quad\Leftrightarrow\quad \begin{bmatrix}I & (I+A+B)^{-1}A \\ A(I+A+B)^{-1} & I \end{bmatrix} \ge 0. \end{eqnarray*} The latter inequality amounts to showing $I \ge (I+A+B)^{-1}A^2(I+A+B)^{-1}$, or in other words, we must show that $(I+A+B)^2 \ge A^2$. Since the map $x \mapsto x^2$ is not operator monotone, suggets that it should be possible to make this inequality fail.

And indeed, here is an explicit counterexample: \begin{equation*} A = \begin{bmatrix}1 &5\\ 5 &61 \end{bmatrix},\quad B = \begin{bmatrix}9 &-9\\ -9 & 13\end{bmatrix},\quad C = (I+A+B)^2-A^2 = \begin{bmatrix}111 &-654\\ -654& 1895 \end{bmatrix}. \end{equation*} The matrix $C$ has a negative eigenvalue: $1003 - 2\sqrt{305845} < 0$.