As was discussed in the comments, it suffices to see that $\delta$ determines $\tau$ uniquely on $\mathrm{dom}(\delta)$ which is dense in $\mathcal{A}$.  Suppose that $x_0 \in \mathrm{dom}(\delta)$.  Check that $t \mapsto \tau^t(x_0)$ is a solution to the initial value problem
\begin{align*}
\frac{d}{dt} x(t)  = \delta( x(t)) && x(0) = x_0.
\end{align*}
If $x$ is any solution then
$$\frac{d}{dt} \left( \tau^{-t}(x(t)) \right) = -\tau^{-t}\delta(x(t)) + \tau^{-t}\left(\delta(x(t))\right) = 0$$
and it follows that $\tau^{-t}(x(t)) \equiv x_0$ so that $x(t) = \tau^t (x_0)$ is the unique solution.

Basically this all works for Banach space flows too. See the answers to my [own question](https://mathoverflow.net/questions/156149/does-the-generator-of-a-1-parameter-group-of-banach-space-isometries-know-which) here.