= Solution
\b[The displayed identity requires a correction to the process construction in the source.] In a <piecewise-deterministic Markov process>, the instantaneous <hazard function> is evaluated at the current flowing state $X_{u-}$, and the destination is sampled from $K_u(X_{u-},dy)$. Keeping the jump times and destinations of the original <Markov jump process> $Y$ instead evaluates both quantities at its last reset state $Y_{u-}$, which generally differs from $X_{u-}$. The source PDF confirms that these original jump times and destinations were retained. Consequently its definition does not in general produce the asserted <transition kernel>, or even a <Markov process> in the state variable $X$ alone.
For an explicit counterexample, take $a(t,x)=1$, $\alpha(x)=2+\tanh x$, $y_0=0$, and let every $K_t(x,dy)$ be the <standard normal distribution>, independently of $t,x$. These are smooth coefficients with bounded positive jump rate, so <nonexplosion> holds. In the literal construction, the first holding time has the <exponential distribution> of rate $2$. Thus the <probability distribution> of $X_t$ has an atom of mass $e^{-2t}$ at $t$. Paths with at least one jump contribute no atom there, because the reset has a <normal density>. The asserted identity instead assigns to this no-jump atom the mass
$$
\exp\left(-\int_0^t(2+\tanh u)\,du\right)=\frac{e^{-2t}}{\cosh t},
$$
which differs for every $t>0$.
Here is the full derivation for the intended <piecewise-deterministic Markov process>. Assume <nonexplosion>; smoothness alone does not guarantee it. Start at $(s,x)$, let $\tau$ be the first jump after $s$, and set
$$
z(u)=\varphi(s,x,u),\qquad H(u)=\int_s^u\alpha(z(r))\,dr.
$$
One precise construction draws an independent unit-rate <exponential random variable> $E$ and takes $\tau=\inf\{u\geq s:H(u)\geq E\}$, with $\inf\varnothing=\infty$. This implements a <hazard function> along the <flow of a vector field>, rather than a fixed exponential holding time at the last reset. Its <survival probability> and jump-time <probability density function> are
$$
S(u)=\mathbb P_{s,x}(\tau>u)=e^{-H(u)},\qquad
-\partial_u S(u)=\alpha(z(u))e^{-H(u)}.
$$
Equivalently, $S'=-\alpha(z)S$ with $S(s)=1$. On $\{\tau>t\}$, the terminal state is $z(t)$. Conditional on $\tau=u$, draw the new state $y$ from $K_u(z(u),dy)$ and restart the same construction from $(u,y)$. The <Markov property> and the <tower property of conditional expectation> therefore give the <first-jump decomposition of a piecewise-deterministic Markov process>
$$
\boxed{\begin{aligned}
(P_{s,t}f)(x)
&=e^{-H(t)}f(z(t))\\
&\quad+\int_s^t\alpha(z(u))e^{-H(u)}
\int_{\mathbb R}(P_{u,t}f)(y)K_u(z(u),dy)\,du.
\end{aligned}}
$$
This proves the requested identity under the corrected construction, for every bounded <Borel measurable function> $f$. The two contributions partition according to whether the first jump occurs before $t$; later jumps are already contained in $P_{u,t}$.
For comparison, the literal construction does have a different renewal identity when started at a reset with $X_s=Y_s=x$. If $Q_{s,t}$ denotes only these reset-start operators, the same <conditional expectation> argument gives
$$
(Q_{s,t}f)(x)=e^{-\alpha(x)(t-s)}f(\varphi(s,x,t))
+\int_s^t\alpha(x)e^{-\alpha(x)(u-s)}(K_uQ_{u,t}f)(x)\,du.
$$
These $Q_{s,t}$ need not be the <transition operators> conditional on $X_s=x$ at an arbitrary time: that observation does not specify the hidden reset state. This distinction explains exactly why replacing only the between-jump paths is insufficient.
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