Compute the Feller Semi-Group of the solution of an ODE. For a homogenous Markov process, we can consider its Feller semi-group: for bounded $f$, it is defined by $$p^{t}f(x)=\mathbb{E}[f(X_{t})|X_{0}=x]=\int_{\mathbb{R}}\mathbb{P}(t,x,dy)f(y),$$ where $\mathbb{P}(t,x,dy)$ is its transition probability density.
Now, For $t\in\mathbb{R}_{\geq 0}$, consider the random ODE $$\frac{dZ_{t}}{dt}=\mathbb{1}_{\{Z_{t}\geq 0\}}.$$ Let $X_{t}$ be a solution of this random ODE, and I want to compute the Feller semi-group of it.

The answer only said $$p^{t}f(x)=\left\{
\begin{array}{ll}
f(x),\ \ \ x<0\\
f(x+t),\ \ \ x\geq 0,
\end{array}
\right.$$
  without further details. 

My attempt is, for each fixed $\omega\in \Omega$, write $$X_{t}(\omega)=\int_{0}^{\infty}\mathbb{1}_{\{X_{t}(\omega)\geq 0\}}dt,$$ then $$p^{t}f(x)=\mathbb{E}\Big[f\Big(\int_{0}^{\infty}\mathbb{1}_{\{X_{t}(\omega)\geq 0\}}dt\Big)|X_{0}=x\Big],$$ then I got stuck....
Is there any alternative way to consider this exercise? Also, why is $X_{t}$ a homogenous Markov process in the first place?
Is there any way to see this process without solving the ODE? or the ODE is easy to solve?
Thank you!
Edit 1:
Thanks to the accepted answer by Michh, I can now compute the Feller semi-group and argue why $X_{t}$ is a homogenous Markov process.
Please see my answer below. I also pointed out that $X_{t}$ does not have Feller property, and gave the transition probability. 
I really appreciate the brilliant answer from Michh, please give him upvote :)
 A: The ODE is easy to solve. Let's start by adding the initial condition: take $Z$ a solution to
$$\frac{d Z_t}{dt} = \mathbf{1}_{Z_t\geq 0}, \quad Z_0 = x.$$
Notice that $\frac{dZ_t}{dt} \geq 0$ so $t\mapsto Z_t$ is nondecreasing. In particular, if $Z_0 = x \geq 0$, we would have $Z_t \geq Z_0 \geq 0$ so that the ODE reduces to $\frac{dZ_t}{dt} = 1$, and $Z_t = x + t$. On the other hand, if $x < 0$, then $Z_t < 0$ for every $t$. To see this, assume by contradiction that there is $t >0$ such that $Z_t \geq 0$ and let $T = \inf\lbrace t >0 \colon Z_t =0\rbrace$ the first time $Z$ becomes positive. Then on $[0,T)$, we would have $Z_t <0$ implying that the RHS of the ODE vanishes and $\frac{dZ_t}{dt} = 0$ for all $t \in [0,T)$. It follows that $Z_t = x<0$ for every $t \in [0,T)$ and $Z_T=0$ is impossible by continuity of $Z$. This proves that $Z_t <0$ for every $t >0$ so that $\frac{dZ_t}{dt} = 0$ and $Z_t = x$ for every $t$.
In conclusion, the solution is given by
$$Z_t = \begin{cases} t+x & \text{if }x \geq 0,\\
x & \text{if } x < 0.
\end{cases}$$
I'll let you compute the Feller semigroup.
A: Thanks to the accepted answer by Michh, I can now answer this post by myself to post how to compute the Feller semi-group and why the solution $X_{t}$ is a homogenous Markov Process.
Firstly, the ODE and the solution being considered in the answer of Michh suggests that $f(X_{t})$ given $X_{0}=x$ is $f(x+t)$ for $x\geq 0$ and $f(x)$ for $x<0$. 
This immediately implies that $$p^{t}f(x)=\mathbb{E}[f(X_{t})|X_{0}=x]=\left\{
   \begin{array}{ll}
   \mathbb{E}f(x+t),\ \ \ x\geq 0\\
   \\
   \mathbb{E}f(x),\ \ \ x<0.
   \end{array}
  \right.
  =\left\{\begin{array}{ll}
   f(x+t),\ \ \ x\geq 0\\
   \\
   f(x),\ \ \ x<0.
   \end{array}
  \right.
$$
Thus, $p^{t}f(x)$ may be discontinuous at $x$. Thus, $p^{t}f(x)$ is not continuous for all bounded continuous function $f$, hence $X_{t}$ does not have Feller property.

In addition, this $p^{t}f(x)$ also suggests the transition probability is of the form $$\mathbb{P}(t,x,\Gamma)=
\left\{
  \begin{array}{ll}
  \delta_{x},\ \ \ x<0\\
  \delta_{x+t},\ \ \ x\geq 0.
  \end{array}
 \right.$$

Finally, to check $X_{t}$ is homogenous Markov, we need to check $\mathbb{P}(s,x,t,\Gamma)=\mathbb{P}(t-s,x,\Gamma)$. But LHS is equivalent to solving an ODE of time index $t\geq s$ with initial condition $X_{s}=x$, which is the same as solving the ODE of time index $t-s\geq 0$ with initial condition $X_{0}=x$ (which is RHS).
It then follows that $X_{t}$ must be Markov since the solution of above ODE is deterministic. 
