Let $(T(t))_{t\geq 0}$ be a strongly continuous semigroup on a Banach space $X$, for which we assume that its growth bound $\omega_0$ is negative. Let $(A,D(A))$ be the generator of $(T(t))_{t\geq 0}.$
$\textbf{Question:}$ How does it implies that $A$ is invertible and $A^{-1}\in\mathfrak{L}(X)$ ?
I assume that you mean $\|T(t)\| \le Me^{\omega_{0}t}$, where $\omega_{0} < 0$. Then $$ Bx=-\int_{0}^{\infty}T(t)x\,dt,\;\;\; x \in X $$ defines a bounded linear operator on $X$ with $\|B\| \le M\frac{1}{|w_{0}|}$. And, $$ BAx = -\int_{0}^{\infty}T'(t)x\,dt = x,\;\;\; x \in \mathcal{D}(A). $$ Similarly, $$ (T(h)-I)Bx = -\int_{0}^{\infty}(T(t+h)-T(t))x\,dt = \int_{0}^{h}T(t)x\,dt. $$ So the following limit exists for all $x \in X$: $$ \lim_{h\downarrow 0}\frac{1}{h}(T(h)-I)Bx = T(0)x = x. $$ So, by the definition of the generator $A$, we have $Bx \in \mathcal{D}(A)$ for all $x$ and $ABx = x$. So $A^{-1}$ exists and equals $B \in \mathcal{L}(X)$.