Does the generator of a 1-parameter group of Banach space isometries know which elements are entire?

If I am not mistaken, then this is connected to the notion of analytic vectors and related object.

If I understand your question correctly, it is answered in this paper of Chernoff.

Further, for group generators on Banach spaces the analytic vectors are dense and they determine the generator $D$, see Exercise II.3.12(2) in Engel-Nagel.

It is well known in the theory of operator semigroups that the Cauchy problem for the generator $D$ of a strongly continuous SEMIgroup is always UNIQUELY solvable for initial data in the domain of $D$, that is $[0,\infty)\to X$, $t\mapsto \sigma_t(x)$ is the unique solution of $f'(t)=D(f(t))$ for all $t\ge 0$ and $f(0)=x$. Therefore, if you find by some other means a solution it coincides with $t\mapsto \sigma_t(x)$. This can be found in any book on operator semigroups, i.e. A short course... from K.-J. Engel and R. Nagel. It should be no problem to extend the result to groups of operators (e.g. by considering $\mathbb R_+$ and $\mathbb R_{-}$ separately).

As it turns out, the basic idea is so simple that I had really might as well add a brief summary. As before, $(\sigma_t)$ is a strongly-continuous, isometric flow on a Banach space $X$ and $D$ is its infinitesimal generator. As I mentioned above, on the face of it the domain of $D$ is just the $x$ for which $\frac{d}{dt} \sigma_t(x) |_{t =0}$ exists, but it follows easily from the group law that, if $x \in \operatorname{dom}(D)$, then $t \mapsto \sigma_t(x)$ is $C^1$ and $\frac{d}{dt} \sigma_t(x) = \sigma_t(D(x))$ is satisfied.

Claim: If $f : \mathbb{R} \to \operatorname{dom}(D)$ is a $C^1$ solution to the initial value problem \begin{align*} \frac{d}{dt} f(t) = D(f(t)) && f(0) = x_0 \in \operatorname{dom}(D), \end{align*} then $f(t) = \sigma_t(x_0)$ for all $t \in \mathbb{R}$.

Proof: Since $$ \frac{d}{dt} \sigma_{-t}( f(t)) = - \sigma_t(D(f(t))) + \sigma_t(\frac{d}{dt}f(t))=0,$$ we conclude $\sigma_{-t}(f(t))$ is constant, hence $\sigma_{-t}(f(t)) = \sigma_0(f(0)) = x_0$ for all $t \in \mathbb{R}$. Applying $\sigma_t$ on both sides gives $f(t) = \sigma_t(x_0)$, as desired.

The above claim shows that $\sigma_t(x)$ is determined by $D$ when $x \in \operatorname{dom}(D)$. In particular, this is true when $x$ is entire, which answers my question. Moreover, since the domain of $D$ is norm-dense in $X$, this implies the flow can be recovered from its infinitesimal generator.

Note the proof of the claim is really no different from the standard proof in elementary, single-variable calculus that, if $\frac{dx}{dt} = \delta x(t)$ and $x(0) = x_0$, then $x(t)=x_0e^{ \delta t}$.

Finally, in the Engel and Nagel book which the other answers mention, the IVP in the claim is replaced by the integral equation $$ f(t) = x_0 + D \left( \int_0^t f(s) \ ds \right).$$ Continuous functions $f : \mathbb{R} \to X$ satisfying this equation are called "mild solutions". In the same way, one can check that $t \mapsto \sigma_t(x_0)$ is the unique mild solution. An advantage to this appraoch is that, because of the extra smoothing afforded by integrating before applying $D$, this actually works for every $x_0 \in X$, and not just for $x_0 \in \operatorname{dom}(D)$ as with the IVP.