No, in general $I-S$ will not be injective.
Indeed, suppose e.g. that $E=\mathbb R^2$ and $$T=\begin{bmatrix}2&1\\ -1&0 \end{bmatrix};$$ here we will identify the linear operators with their matrices in the standard basis of $\mathbb R^2$.
Then (assuming $G$ and $L$ are the orthogonal complements of $N(I-T)$ and $R(I-T)$ respectively) we have $$\pi_1=\frac12\begin{bmatrix}1&-1\\ -1&1 \end{bmatrix};$$ $$\pi_2=\frac12\begin{bmatrix}1&1\\ 1&1 \end{bmatrix};$$ $$\Lambda=\begin{bmatrix}0&0\\ 0&0 \end{bmatrix};$$ $$I-S=I-T=\begin{bmatrix}-1&-1\\ 1&1 \end{bmatrix}.$$ So, $I-S$ is not injective.