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 here.

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 here.

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 [question](I asked a related question here.

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 here.