Let
- $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space and $$\mathcal N:=\left\{N\in\mathcal A:\operatorname P[N]=0\right\}$$
- $(W_t)_{t\ge0}$ be a Brownian motion on $(\Omega,\mathcal A,\operatorname P)$ and $$W^{(s)}_t:=W_{s+t}-W_s\;\;\;\text{for }t\ge0$$ for $s\ge0$
- $b,\sigma:[0,\infty)\times\mathbb R\to\mathbb R$ be Borel measurable with $$|b(t,x)|^2+|\sigma(t,x)|^2\le K(1+|x|^2)\;\;\;\text{for all }x\in\mathbb R\tag1$$ and $$|b(x)-b(y)|^2+|\sigma(x)-\sigma(y)|^2\le K|x-y|^2\;\;\;\text{for all }x,y\in\mathbb R\tag2$$ for some $K\ge0$
We know that for all $s\ge0$ there is a real-valued continuous process $\left(X^{(s,\:x)}_t\right)_{(t,\:x)\in[s,\:\infty)\times\mathbb R}$ on $(\Omega,\mathcal A,\operatorname P)$ with $$X^{(s,\:x)}_t=x+\int_s^tb\left(X^{(s,\:x)}_r\right)\:{\rm d}r+\int_s^t\sigma\left(X^{(s,\:x)}_r\right)\:{\rm d}W_r\tag3$$ for all $t\ge s$ almost surely for all $x\in\mathbb R$.
Now, let $\xi$ be a real-valued random variable on $(\Omega,\mathcal A,\operatorname P)$ independent of $W$ and $$\mathcal F_t:=\sigma(\xi)\vee\sigma(W_s:s\in[0,t])\vee\mathcal N\;\;\;\text{for }t\ge0.$$ Now, there is a real-valued $\mathcal F$-adapted continuous process $(X_t)_{t\ge0}$ with $$X_t=\xi+\int_0^tb(X_s)\:{\rm d}s+\int_0^t\sigma(X_s)\:{\rm d}W_s\;\;\;\text{for }t\ge0.$$ $X$ is a $\mathcal F$-Markov process with $$\operatorname P\left[X_t\in B\mid\mathcal F_s\right]=\kappa_{s,\:t}(X_s,B)\;\;\;\text{almost surely for all }t\ge0\text{ and }B\in\mathcal B(\mathbb R),\tag4$$ where $$\kappa_{s,\:t}(x,B):=\operatorname P\left[X^{(s,\:x)}_t\in B\right]\;\;\;\text{for }t\ge s\ge0\text{ and }(x,B)\in\mathbb R\times\mathcal B(\mathbb R).$$
How can we show that the transition semigroup $(\kappa_{s,\:t}:t\ge s\ge0)$ of $X$ is time-homogeneous, i.e. that $\kappa_{s,\:t}$ only depends on $t-s$ for all $t\ge s\ge0$?
Clearly, we may observe that $$X^{(0,\:x)}_h=x+\int_0^hb\left(X^{(0,\:x)}_r\right)\:{\rm d}r+\int_0^h\sigma\left(X^{(0,\:x)}_r\right)\:{\rm d}W_r\tag5$$ for all $h\ge0$ almost surely for all $x\in\mathbb R$, while $$X^{(s,\:x)}_{s+h}=x+\int_0^hb\left(X^{(s,\:x)}_{s+r}\right)\:{\rm d}r+\int_0^h\sigma\left(X^{(s,\:x)}_{s+r}\right)\:{\rm d}W_r^{(s)}\tag6$$ for all $h\ge0$ almost surely for all $s\ge0$ and $x\in\mathbb R$.
My guess is that $(5)$ and $(6)$ imply that $\left(X^{(0,\:x)}_h\right)_{h\ge0}$ and $\left(X^{(s,\:x)}_{s+h}\right)_{h\ge0}$ have the same distribution (since $W$ and $W^{(s)}$ have the same distribution) for all $s\ge0$ and $x\in\mathbb R$. How can we show that rigorously and how can we conclude from that?