Let $\Omega\subseteq\mathbb{C}$ be open and let $B(X)$ be the Banach algebra of all bounded linear operators on the Banach space $X$. Let $f_{\alpha}:\Omega\to\mathbb{C}$ be such that $f_{\alpha}(z)=e^{-\alpha z}$ for $\alpha\in\mathbb{R}$ and define the family $$e^{-\alpha T}=f_{\alpha}(T).$$
I want to prove the group property in the title.
So far, I have said: let $A:=B(X)$, and let $H(\Omega)$ be the algebra of all holomorphic functions in $\Omega$. Define $A_{\Omega}:=\{T\in B(X):\sigma(T)\subset\Omega\}$ be an open subset of $A$. Note that $\exp\in H(\Omega)$.
Now define $\widetilde{H}(A_{\Omega})$ as the set of all A-valued functions $\widetilde{f}$ with domain $A_{\Omega}$ that arise from a $f\in H(\Omega)$ by the formula $$\widetilde{f}(x)=\frac{1}{2\pi i}\int_{\Gamma}f(\lambda)(\lambda e-x)^{-1}d\lambda,$$ where $\Gamma$ is any contour that surrounds $\sigma(T)$ in $\Omega$. Note also that $\widetilde{H}(A_{\Omega})$ is a complex algebra and the mapping $f\to\widetilde{f}$ is an algebra isomorphism of $H(\Omega)$ onto $\widetilde{H}(A_{\Omega})$.
Now, the resolvent for $|\lambda|>\|T\|$ is $$(\lambda e-T)^{-1}=\lambda^{-1}\left(e-\frac{1}{\lambda}T\right)^{-1}=\frac{1}{\lambda}\sum_{n=0}^{\infty}\frac{1}{\lambda^{n}}T^{n}=\sum_{n=0}^{\infty}\frac{1}{\lambda^{n+1}}T^{n},$$ and so integrating around a contour that encloses $|\lambda|\le\|T\|$ gives that $$\widetilde{\exp}(T)=\frac{1}{2\pi i}\int_{\Gamma}e^{\lambda}\sum_{n=0}^{\infty}\frac{1}{\lambda^{n+1}}T^{n}d\lambda=\sum_{n=0}^{\infty}\left(\frac{1}{2\pi i}\int_{\Gamma}\frac{e^{\lambda}}{\lambda^{n+1}}d\lambda\right)T^{n}=\sum_{n=0}^{\infty}\frac{1}{n!}T^{n}.$$ Since $\exp\in H(\Omega)$, we can write $$e^{-(\alpha+\beta)z}=e^{-\alpha z}e^{-\beta z},\qquad (z\in \Omega)$$ I don't really know where to go from here though. I think a lot of this might have also been unnecessary.
