Semi-group property of branching processes? See the edit below!
I do have a question about continuous-state branching processes while reading "Fluctuations of Lévy Processes with Applications" by Kyprianou:

Definition (Continuous-state branching process).
A $[0, \infty]$-valued strong Markov process $Y=\{Y_t : t \geq 0\}$ with probabilities $ \{ P_x : x \geq 0 \}$ is called a continuous-state branching process if it has càdlàg paths and satisfies the branching property.

Let $Y$ be a continuous-state branching process. Then it is shown that the semi-group property
$$ 
   u_{t+s}(\theta) = u_t(u_s(\theta))
$$
is true where $u_r(\theta)$ fulfills the following equation:
$$
   E_x(e^{- \theta Y_r}) = e^{-xu_r(\theta)} \tag{$\star$} .
$$
One step in the proof states
$$
  E_x(E(e^{-\theta Y_{t+s}}|Y_t)) = E_x(e^{-Y_t u_s(\theta)}).
$$
I don't understand this step. What I tried to do was the following:
$$
E(e^{-\theta Y_{t+s}}|Y_t) = E(e^{-\theta (Y_{t+s} - Y_t)} e^{-\theta Y_t}|Y_t) = E(e^{-\theta (Y_{t+s} - Y_t)}|Y_t) e^{-\theta Y_t} = E_{K}(e^{-\theta (\tilde{Y}_s - K)}|Y_t) e^{-\theta Y_t}
$$
Here, I used in the second equality measurability. In the third equality, I want to use something of the form that $Y_{t+s} - Y_t$ is distributed like $\tilde{Y}_s$. But I don't want to start $\tilde{Y}$ at zero since  such a branching process would then always be zero. As an alternative I thought of something like $\tilde{Y}_s - K$ where $\tilde{Y}$ is started at $K = Y_t$. However, I guess I am doing something wrong since I think we don't have any time homogeneity that I used and furthermore what I am doing with $K$ feels also weird but I don't know what else I should be doing since the process $Y_s$ is always zero if it is started at zero.
Assuming, however this is correct, I do can continue in the following way:
$$
E_{K}(e^{-\theta (\tilde{Y}_s - K)}|Y_t) e^{-\theta Y_t} = E_{Y_t}(e^{-\theta \tilde{Y}_s}|Y_t) e^{\theta K} e^{-\theta Y_t} = e^{-\tilde{Y}_t u_s(\theta)}
$$
which shows the desired step.
So my question is how this step is done properly and maybe also what my misconceptions are?
Thanks a lot for your help.
Edit:
I was told that from the Markov property we can follow the equation
$$
E(e^{-\theta Y_{t+s}}|Y_t) = E_{Y_t} (e^{-\theta Y_s})
$$
and using $(\star)$ the desired result follows.
However, I don't understand this step with the Markov property formally.
Intuitively, it makes sense that the equation is true since if we know $Y_t$ (or $\mathcal{F}_t$ by the Markov property), the process has only time s to evolve from this, so this describes the same situation when would start a process at $Y_t$ and let it again evolve for time s. However, this sounds to me like some time homogeneity is needed and I don't see how this follows formally from the (strong) Markov property, i.e.:
$$
P(Y_{t+s} \in B | \mathcal{F}_t) = P(Y_{t+s} \in B | Y_t).
$$
So, I am wondering what I am missing/don't understand correctly in this situation.
Thank you.
 A: If $(X_t)_{t \geq 0}$ is any Markov process, then the very definition of the Markov property is that
\begin{align} \tag{0}
E_x[f(X_{t+s})|X_t] = E_{X_t}[f(X'_s)].
\end{align}
where $(X'_s)_{s \geq 0}$ is a copy of the Markov process starting from $X_t$ under $E_{X_t}$.

We will use both the Markov property and the branching property (i.e. $E_x[e^{-\theta Y_t} ] = e^{ - xu_t(\theta)}$) to prove that $u$ satisfies the semigroup property.
As you observe, by the tower property we have
\begin{equation} \tag{1}
E_x[e^{-\theta Y_{t+s}} ] = E_x[E_x[e^{-\theta Y_{t+s} } | Y_t]]
\end{equation}
By the Markov property,
\begin{equation} \tag{2}
E_x[e^{-\theta Y_{t+s} } | Y_t] = E_{Y_t}[e^{ - \theta Y'_s}],
\end{equation}
where $(Y'_s)_{s \geq 0}$ is an independent copy of the CSBP. Both sides of (2) refer to $\mathcal{F}_t$-measurable random variables, where $(\mathcal{F}_s)_{s \geq 0}$ is the underlying filtration. ((2) is the special case of (0) for $f(y) = e^{ - \theta y}$.)
Using the definition of $u_s(\theta)$, we have $E_{Y_t}[e^{ - \theta Y'_s}] = e^{ - Y_t u_s(\theta)}$, so that plugging this fact into (2) we obtain
\begin{equation} \tag{3}
E_x[e^{-\theta Y_{t+s} } | Y_t] = e^{-Y_tu_s(\theta)}.
\end{equation}
Plugging (3) into (1) we have
\begin{equation} \tag{4}
E_x[e^{-\theta Y_{t+s}} ] = E_x[e^{-Y_tu_s(\theta)}].
\end{equation}
Writing $\lambda := u_s(\theta)$, the right-hand-side of (4) is $E_x[e^{-\lambda Y_t}]$, which is by definition of $u_t(\lambda)$ equal to $e^{ - x u_t(\lambda)}$. Substituting in $\lambda := u_s(\theta)$ we finally obtain
\begin{equation} \tag{4}
E_x[e^{-\theta Y_{t+s}} ] = e^{ - xu_s(u_t(\theta))},
\end{equation}
as required.
