Suppose $\im L$ is closed. Use the uniform boundedness principle to show that the family of operators $tA_t$, $t > 0$ is uniformly bounded on $\im L$, i.e. there is some constant $C < \infty$ such that for all $x\in\im L$:
$$
\sup\{\norm{tA_tx}: t > 0\}\leq C\Vert x\Vert~.
$$
2. If moreover $E=\ker L\oplus\im L$, then $\lim_{t\to\infty}\norm{A_t-Q}=0$ and $\lim_{\l\to0}\norm{\l(\l-L)^{-1}-Q}=0$.
For $x=Ly$ we have
$$
tA_tx=\int_0^tP_sLy\,ds=P_ty-y
$$
it follows that for all $x\in\im L$: $\sup\{\norm{tA_tx}:t > 0\}\leq2\norm y < \infty$. By the uniform boundedness principle we conclude that $\sup\{\norm{tA_t|\im L}:t > 0\} < \infty$. The same argument applies to the family $\l(\l-L)^{-1}$, $\l > 0$: For $x=Ly$:
$$
\norm{(\l-L)^{-1}x}
=\norm{-y+\l(\l-L)^{-1}y}
\leq2\norm y~.
$$
2. For $x\in E$ we have the decomposition $x=Qx+(1-Q)x$. Hence
$$
\norm{A_tx-Qx}
=\norm{A_t(1-Q)x}
\leq Ct^{-1}\norm{(1-Q)x}
\leq2Ct^{-1}\Vert x\Vert~.
$$