No, in general $M:=I-S$ 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 
$$\pi_1=\frac12\begin{bmatrix}1&-1\\ -1&1 \end{bmatrix};$$