The generator of the contour semigroup is the sectorial operator

The inclusion $A\subseteq B$. First let $y\in D(A)$ and $s>0$. The contour formula for $T(s)Ay$ and [L1] give $T(s)Ay=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}{e}^{\lambda s}R(\lambda ,A)Ayd\lambda =\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}{e}^{\lambda s}(\lambda R(\lambda ,A)y-y)d\lambda =T^{\mathrm{\prime }}(s)y=AT(s)y\mathrm{.}$ Here $R(\lambda ,A)Ay=\lambda R(\lambda ,A)y-y$ follows from [L6], and $(2\pi i{)}^{-1}{\int }_{\mathrm{\Gamma }}{e}^{\lambda s}d\lambda =0$: close a truncated keyhole, use the entire primitive $e^{\lambda s}\mathrm{/}s$, and let its exponentially decaying outer arc tend to zero. Now for $x\in D(A)$ and $h>0$, the fundamental theorem [L3] on $[\epsilon ,h]$, followed by $\epsilon \downarrow 0$ using strong continuity at $0$, gives $T(h)x-x={\int }_0^{h}T(s)Axds$. Dividing by $h$ and applying average convergence [L3] yields $h^{-1}(T(h)x-x)\to Ax$. By the generator definition in [L7], $x\in D(B)$ and $Bx=Ax$.

The resolvent computation. Fix $x\in X$ and $\lambda >\max \{\omega ,0\}$, and put $v_{\eta ,R}:={\int }_{\eta }^R{e}^{-\lambda t}T(t)xdt$ for $0<\eta <R$. On the compact interval $[\eta ,R]$, both $t\mapsto T(t)x$ and $t\mapsto AT(t)x=T^{\mathrm{\prime }}(t)x$ are continuous, so the orbit is continuous in the graph norm of the closed operator $A$. Its graph-norm Bochner integral therefore lies in $D(A)$ and satisfies $A{v}_{\eta ,R}={\int }_{\eta }^R{e}^{-\lambda t}AT(t)xdt$. Since $(d\mathrm{/}dt)[e^{-\lambda t}T(t)x]=e^{-\lambda t}(AT(t)x-\lambda T(t)x)$, the fundamental theorem [L3] gives $(\lambda I-A)v_{\eta ,R}={\int }_{\eta }^R{e}^{-\lambda t}(\lambda T(t)x-AT(t)x)dt=-{\int }_{\eta }^R\frac{d}{dt}(e^{-\lambda t}T(t)x)dt=e^{-\lambda \eta }T(\eta )x-e^{-\lambda R}T(R)x$. As $\eta \downarrow 0$ and $R\to \mathrm{\infty }$, $v_{\eta ,R}\to v:={\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}T(t)xdt$ by the exponential bound and strong continuity at $0$, while the displayed right side tends to $x$ because $\lambda >0$. Closedness of $A$ now gives $v\in D(A)$ and $(\lambda I-A)v=x$; since $\lambda >0$ and $A$ is sectorial, $\lambda \in \rho (A)$ and $v=R(\lambda ,A)x$.

The resolvents agree and $B=A$. By [step 1.2] the vector $v={\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}T(t)xdt$ equals $R(\lambda ,A)x$; by the Laplace formula [L2] the same integral equals $R(\lambda ,B)x$. Hence $R(\lambda ,B)x=R(\lambda ,A)x$ for every $x\in X$. Their ranges agree and equal $D(B)=D(A)$; for each $y$ in this common domain, applying the inverses gives $(\lambda I-B)y=(\lambda I-A)y$, so $B=A$.

Uniqueness among exponentially bounded semigroups. Let $S$ and $T$ be two strongly continuous semigroups with generator $A$ and exponential bounds, fix $t>0$ and $x\in D(A)$, and set $g(s):=S(t-s)T(s)x$ for $s\in [0,t]$. For $0<s<t$, write the difference quotient as $\frac{g(s+h)-g(s)}{h}=\frac{S(t-s-h)-S(t-s)}{h}T(s)x+S(t-s-h)\frac{T(s+h)x-T(s)x}{h}\mathrm{.}$ By [L5], $T(s)x\in D(A)$, the first term tends to $-S(t-s)AT(s)x$, and the second tends to $S(t-s)AT(s)x$; hence $g^{\mathrm{\prime }}(s)=0$. Continuity at the endpoints and the fundamental theorem [L3] show that $g$ is constant, so $S(t)x=T(t)x$. Since $D(A)$ is dense by [L4] and $S(t),T(t)$ are bounded, this extends to all $x\in X$.

Assembly. [step 2.1] identifies the generator of the contour semigroup with $A$, and [step 2.2] proves uniqueness in the exponentially bounded class; no choice principle beyond Dependent Choice was used, since only the contour construction, the Laplace representation and the fundamental theorem were invoked.