My question is what's written in the title, that is, if $T$ is a normal operator on a Hilbert space $H$, and $T$ is left invertible, is it necessarily true that $T$ is invertible?
Actually, the more general question I have is if $T$ is normal and $T-\lambda I$ ($\lambda\in\mathbb{C}$) is left invertible, is it true that $T-\lambda I$ is invertible?
The second question is the same as the first one.
Let $S$ be a left inverse for $T$. The existence of $S$ implies that $\ker T=\{0\}$. For any $x\in H$, $$ \|x\|=\|STx\|\leq\|S\|\,\|Tx\|=\|S\|\,\|T^*x\| $$ (this is where we use that $T$ is normal). So $T^*$ is bounded below. This implies that $T^*$ is one-to-one and that its range is closed. Also $$ \text{ran}\,T^*=\overline{\text{ran}\,T^*}=(\ker T)^\perp=\{0\}^\perp=H. $$ So $T^*$ is bijective, which implies (by the Open Mapping Theorem) that it is invertible. And so $T$ is also invertible.