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](http://mathoverflow.net/questions/156149/does-the-generator-of-a-1-parameter-group-of-banach-space-isometries-know-which) here.