Is the product rule true in a Banach algebra?

Yes, you can. For $h \ne 0$, $$ \begin{align} & \frac{1}{h}\left[T(t+h)F(t+h)-T(t)F(t)\right]\\ & =\frac{1}{h}\{T(t+h)-T(t)\}F(t+h)+T(t)\frac{1}{h}\{F(t+h)-F(t)\}\\ & =\frac{1}{h}\{T(t+h)-T(t)\}F(t) + T(t)\frac{1}{h}\{F(t+h)-F(t)\} \\ & + \frac{1}{h}\{T(t+h)-T(t)\}\{F(t+h)-F(t)\} \end{align} $$ Therefore, $$ \begin{align} & \left\|\frac{1}{h}\left[T(t+h)F(t+h)-T(t)F(t)\right]-\left[T'(t)F(t)+T(t)F'(t)\right]\right\| \\ & \le \left\|\frac{1}{h}[T(t+h)-T(t)]-T'(t)\right\|\|F(t)\| +\|T(t)\|\left\|\frac{1}{h}[F(t+h)-F(t)]-F'(t)\right\| \\ & + \left\|\frac{1}{h}[T(t+h)-T(t)]\right\|\left\|\frac{1}{h}[F(t+h)-F(t)]\right\||h|. \end{align} $$ The right side tends to 0 as $h\rightarrow 0$ assuming that $T$ and $F$ are differentiable at $t$.