Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(E,\mathcal E)$ be a measurable space and $X:\Omega\times[0,\infty)\times E\to E$ be a stochastic flow, i.e. $X$ is $(\mathcal A\otimes\mathcal B([0,\infty))\otimes\mathcal E,\mathcal E)$-measurable and $X(\omega,\;\cdot\;,\;\cdot\;)$ is a flow for all $\omega\in\Omega$. Now let $$X^x_t:=X(\;\cdot\;,t,x)\;\;\;\text{for }(t,x)\in[0,\infty)\times E.$$ It's easy to see that $$\kappa_t(x,B):=\operatorname P\left[X^x_t\in B\right]\;\;\;\text{for all }(x,B)\in E\times\mathcal E$$ is a Markov kernel and $$(\kappa_tf)(x):=\int\kappa(x,{\rm d}y)f(y)=\operatorname E\left[f(X^x_t)\right]\tag1$$ for all $\mathcal E$-measurable $f:E\to[0,\infty)$ for all $t\ge0$.
Question 1: Can we show that $(\kappa_t)_{t\ge0}$ is a semigroup, i.e.$^1$ $\kappa_s\kappa_t=\kappa_{s+t}$ for all $s,t\ge0$?
Question 2: Assuming that $E$ is a $\mathbb R$-Banach space, $\mathcal E=\mathcal B(E)$ and $X(\omega,t,\;\cdot\;)$ is continuously Fréchet differentiable for all $(\omega,t)\in\Omega\times[0,\infty)$, are we able to show that $\kappa_tf$ is continuously Fréchet differentiable for all continuously Fréchet differentiable $f:E\to\mathbb R$ and $t\ge0$?
Regarding question 1: Let $s,t\ge0$. By the flow property, we know that $$X^{X^x_s}_t=X^x_{s+t}\;\;\;\text{for all }x\in E\tag2$$ and hence $$(\kappa_s\kappa_t)(x,B)=\operatorname E\left[\kappa_t(X^x_s,B)\right]=\operatorname E\left[\left.\operatorname P\left[X^y_t\in B\right]\right|_{y=X^x_s}\right]\tag3$$ for all $(x,B)\in E\times\mathcal E$.
Regarding question 2: If $f:E\to\mathbb R$ is Fréchet differentiable, then $E\ni x\mapsto f(X^x_t(\omega))$ is clearly Fréchet differentiable by the chain rule for all $(\omega,t)\in\Omega\times[0,\infty)$. So, I guess the desired claim boils down to a suitable application of the dominated convergence theorem.
$^1$ $\kappa_s\kappa_t$ denotes the composition of kernels.