Let $X$ be a Banach space and $T ∈ B(X)$. Let $f_t(ζ) = e^{-tζ}$ , with $f_t : Ω → \mathbb{C}$ and $t ∈ \mathbb{R}$. Define the exponential family $e^{−tT} = f_t(T)$.
Show that $e^{−tT}$ forms a group. That is, $e^{−(t+s)T} = e^{−tT} e^{−sT}$ for all $s, t ∈ \mathbb{R}$.
I don't really understand what to do, actually..
From the way you state you state your problem, it looks like you are thinking of $e^{-tT}$ as being defined by holomorphic functional calculus (By the way, do you want $\Omega = \mathbb{C}$? Because your functions $f_t$ are entire...). If you already understand this tool, then the conclusion you want holds because, at the level of functions, $f_t f_s = f_{t+s}$, and holomorphic functional calculus is a homomorphism, i.e. $(fg)(T) = f(T)g(T)$.
On the other hand, you can also define $e^{-tT}$ using the absolutely convergent series $\sum_{n=0}^\infty (-tT)^n/n!$. If that is the route you are taking, then you should just follow the proof that $e^{s+t} = e^s e^t$ in the single-variable context. It goes through more or less unchanged.