Let $(E,\mathcal E)$ be a measurable space and $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$. Note that $(\kappa_t)_{t\ge0}$ is a contraction semigroup on $$\mathcal E_b:=\left\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\right\}$$ (equipped with the supremum norm) via $$(\kappa_tf)(x):=\int\kappa_t(x,{\rm d}y)f(y)\;\;\;\text{for }f\in\mathcal E_b.$$
Now let $\mu$ be a probability measure on $(E,\mathcal E)$ which is invariant with respect to $(\kappa_t)_{t\ge0}$ and $p\ge1$. Noting that $\mathcal E_b$ is dense in $\mathcal L^p(\mu)$, we easily see that $\kappa_t$ has a unique contractive linear extension to $L^p(\mu)$ for all $t\ge0$ and $(\kappa_t)_{t\ge0}$ is a contraction semigroup on $L^p(\mu)$.
Assuming that $$[0,\infty)\times E\to\mathbb R\;,\;\;\;(t,x)\mapsto(\kappa_tf)(x)\tag1$$ is $\mathcal B([0,\infty))\otimes\mathcal E$-measurable for all $f\in\mathcal E_b$, are we able to show that $(1)$ and/or $$[0,\infty)\to L^p(\mu)\;,\;\;\;f\mapsto\kappa_tf\tag2$$ is $\mathcal B([0,\infty))\otimes\mathcal B(L^p(\mu))$-measurable for all $f\in L^p(\mu)$?
EDIT: We know that there is a nondecreasing $(f_n^\pm)_{n\in\mathbb N}\subseteq\mathcal E_b$ with $f^\pm_n\xrightarrow{n\to\infty}f^\pm$ and hence $$(\kappa_tf^\pm_n)(x)\xrightarrow{n\to\infty}(\kappa_tf^\pm)(x)\;\;\;\text{for all }(t,x)\in[0,\infty)\times E\tag3.$$ Thus, $(1)$ is measurable for $f$ replaced by $f^\pm$.
Now, since $f\in\mathcal L^1(\mu)$, we only obtain that $\mu\kappa_t|f|<\infty$ and hence $$\kappa_t|f|<\infty\;\;\;\mu\text{-almost surely}\tag4$$ for all $t\ge0$. So, $$\kappa_tf=\kappa_tf^+-\kappa_tf^-\tag5$$ does only hold up to a $\mu$-null set depending on $t\ge0$. This should prevent us from concluding the desired measurability of $(1)$, unless we are assuming that $\mathcal A$ is $\mu$-complete (or is even that not enough?).
Regarding the measurability of $(3)$, I have no idea how we could show that ...