In [Li, Xungjing, and Jiongmin Yong. Optimal control theory for infinite dimensional systems.1995]
at page 241 it is claimed that:
\begin{equation}
\lim _{s \downarrow t}\left|e^{A^{*}(s-t)} \frac{y_{t, x}(s)}{\left|y_{t, x}(s)\right|}-\frac{x}{|x|}\right|=0
\end{equation}
uniformly in $u(\cdot) \in \mathcal{U}[t, T]$.
Here the state equation is
$$y_{t, x}(s)=e^{A(s-t)} x+\int_{t}^{s} e^{A(s-r)} f\left(r, y_{t, x}(r), u(r)\right) d r, \quad s \in[t, T], x \in X$$
where $X$ is Hilbert, $A : D(A) \subset X \to X$ linear operator is the generator of a strongly continuous semigroup of contractions $|e^{At}|\leq 1$, $U$ a metric space in which the control $u(\cdot)$ takes values $u(\cdot) \in \mathcal{U}[0, T] \equiv$ $\{u:[0, T] \rightarrow U \mid u(\cdot)$ measurable $\}$, $f : [0,T] \times X \times U \to X$ satisfies
$$|f(t, x, u)-f(\bar{t}, \bar{x}, u)| \leq L|x-\bar{x}|+\omega(|t-\bar{t}|)$$
$\forall t, \bar{t} \in[0, T], x, \bar{x} \in X, u \in U$ and modulus of continuity $\omega$.
$$|f(t, 0, u)| \leq L, \quad \forall(t, u) \in[0, T] \times U$$.
Note that $A^*$ is the adjoint of $A$ and $e^{A^* t}$ is a strongly continuous semigroup of contractions with generator $A^*$.
I don't manage to prove the claim, how do you show that?
As discussed in the comments, $y_{t,x}(s)\to x$ as $s\searrow t$. Thus $y_{t,x}(s)/|y_{t,x}(s)|\to x/|x|$.
Moreover, $(e^{rA^\ast})_{r\geq 0}$ is a strongly continuous semigroup. I would even say this is implicit in the definition, i.e., $(e^{rA^\ast})$ is defined to be the $C_0$-semigroup generated by $A^\ast$. In any case, the fact that the adjoint of a generator of a $C_0$-semigroup on a Hilbert space (or more generally on a reflexive Banach space) is again a generator of a $C_0$-semigroup is discussed in 5.14 of Engel, Nagel: One-Parameter Semigroups for Linear Evolution Equations.
This result relies on two ingredients: First, it is not hard to see that the adjoint of a $C_0$-semigroup is weak$^\ast$-continuous, hence weakly continuous if the underlying Banach space is reflexive. Then, and that's the harder part, one uses the uniform boundedness principle and the Hahn-Banach theorem to show that every weakly continuous semigroup is in fact strongly continuous.
Therefore, $e^{(s-t)A^\ast}\to \mathrm{id}_X$ strongly as $s\searrow t$ and $s\mapsto e^{(s-t)A^\ast}$ is norm bounded on compact intervals. Elementary estimates then show that $$ \lim_{s\searrow t}e^{(s-t) A^\ast}\frac{y_{t,x}(s)}{|y_{t,x}(s)|}= \frac{x}{|x|}. $$