Yes.

I assume that $E$ is a Polish space, or a standard Borel space that can be turned into a Polish space by some auxillary topology.

Basically, you can view your measurable stochastic process as a random element in the space $L^0(\mathbb{R}_+, E)$ of equivalence classes of $E$-valued measurable functions. $L^0(\mathbb{R}_+, E)$ comes equipped with the Polish topology of convergence in measure on $[0,n]$ for all $n \in \mathbb{N}$ and its corresponding standard Borel $\sigma$-algebra. While doing that you lose some information (e.g. the value for some specific $t$), but you will be able to calculate your Lebesgue measures as measurable functions on $L^0$.

Note that this random element in $L^0(\mathbb{R}_+, E)$ doesn't change if you replace the process by a version of itself. Namely, if $X$ and $X^\prime$ are versions of each other then for every measurable set $A \subset E$ the sets $\{t \mid X_t \in A\}$ and $\{t \mid X^\prime_t \in A\}$ will almost surely differ on a set of Lebesgue measure $0$, by Fubini's theorem: $$\mathsf{E} \intop | \mathsf{1}\{X_t \in A\} - \mathsf{1}\{X^\prime_t \in A\} | dt = \intop \mathsf{E} | \mathsf{1}\{X_t \in A\} - \mathsf{1}\{X^\prime_t \in A\} | dt = 0$$

Now your stationarity assumption means that there is a measure-preserving transformation $T_s: \Omega \to \Omega$, such that $(X_{t+s}, t \ge 0)$ and $(X_t \circ T_s, t \ge 0)$ are versions of each other, so the $L^0$-valued random variables corresponding to them will be the same.

Yes.

I assume that $E$ is a Polish space, or a standard Borel space that can be turned into a Polish space by some auxillary topology.

Basically, you can view your measurable stochastic process as a random element in the space $L^0(\mathbb{R}_+, E)$ of equivalence classes of $E$-valued measurable functions. $L^0(\mathbb{R}_+, E)$ comes equipped with the Polish topology of convergence in measure on $[0,n]$ for all $n \in \mathbb{N}$ and its corresponding standard Borel $\sigma$-algebra. While doing that you lose some information (e.g. the value for some specific $t$), but you will be able to calculate your Lebesgue measures as measurable functions on $L^0$.

Now I claim that

The distribution of the $L^0$ version of $X$ can be recovered from the joint distribution of $(X_t, t \ge 0)$.

Note that this will immediately imply a positive answer to your question.

Here is a sketch of a proof of my claim. Consider the functions $L^0(\mathbb{R}_+, E) \to \mathbb{R}$ of the following form:

$$F_{f,T}: x \mapsto \intop_0^T f(x_t) dt,$$

where $f: E \to [0,1]$ is bounded and Borel, and $T \ge 0$. These functions generate the whole Borel $\sigma$-algebra on $L^0$, because the level sets $\{F_{f,T} < \mathrm{const}\}$ for continuous $f: E \to [0,1]$ generate the topology.

Now it remains to pin down the joint distribution of $F_{f,T}(X)$. This is done by observing that the moments can be calculated using the joint distribution of $(X_t)$ by Fubini's theorem:

$$\mathsf{E} \intop_0^T f_1(X_t) dt \dots \intop_0^T f_n(X_t) dt = \intop_{[0,T]^n} \mathsf{E} f_1(X_{t_1}) \dots f_n(X_{t_n}) dt_1 \dots dt_n$$