The problem is locally well-posed, i.e., the problem admits a unique
solution $y\in C([0,T];{\mathcal X})$ for some (in general small) $T>0$. In addition, it holds that $\dot{y}\in L^\infty((0,T);{\mathcal X})$ if $f\in D(A)$.
This can be proven by standard arguments. Indeed, denote the nonlinearity by $F$
(as $f$ is the initial value). There are constants $K\geq1$, $L>0$,
and $T>0$ such that $\|e^{tA}\|_{\mathcal L(\mathcal X)} \leq K e^{K t}$
for all $t\geq0$,
\begin{align*}
& \|f\|_{\mathcal X} + LK\left(K+2\right) T \leq K, \quad \|F(0,0)\|_{\mathcal X}\leq K, \quad
\|u\|_{L^\infty((0,T);{\mathcal X}')}\leq K, \\
& \|F(y,u)-F(\bar y,\bar u)\|_{\mathcal X} \leq L\left(\|y-\bar y\|_{\mathcal X}+\|u-\bar u\|_{\mathcal X'}\right)
\end{align*}
if $\max\{\|y\|_{\mathcal X},\|\bar y\|_{\mathcal X}\}\leq K^2$, $\max\{\|u\|_{{\mathcal X}'},\|\bar u\|_{{\mathcal X}'}\}\leq K$, and $LKT<1$. Define
$$
\mathbb X=\{y\in C([0,T];{\mathcal X})\mid y(0)=f,\,\|e^{-K
t}y\|_{L^\infty((0,T);{\mathcal X})}\leq K^2\}
$$
which is non-empty and a complete metric space furnished with the metric
$(y,\bar y)\mapsto \|e^{-Kt} y-e^{-Kt}\bar
y\|_{L^\infty((0,T);{\mathcal X})}$. Define further
$$
(\Phi y)(t) = e^{tA}f + \int_0^t e^{(t-s)A}F(y(s),u(s))\,ds, \quad 0\leq t\leq T.
$$
Then $\Phi\colon \mathbb X\to\mathbb X$ and $\|e^{-Kt}\Phi y-e^{-Kt}\Phi \bar y\|_{L^\infty((0,T);{\mathcal X})}\leq LKT \, \|e^{-Kt}y-e^{-Kt}\bar y\|_{L^\infty((0,T);{\mathcal X})}$ for $y,\bar y\in\mathbb X$. So the contration mapping principle applies and demonstrates that $\Phi$ has a unique fixed point $y\in\mathbb X$ which is the desired solution.
