I have to show that the following family of kernels is a family of Markovian transition functions on $(\mathbb{R}_+,\mathcal{B}(\mathbb{R}_+))$: \begin{equation}\tag{1}\label{eq:identity} P_t f(x) = e^{-t/x}f(x) + \int_x^{\infty} \frac{t}{y^2} e^{-t/y}f(y) \, \text{d}y. \end{equation}
So I have to prove the Chapman-Kolmogorov equation: \begin{equation}\tag{2}\label{eq:ck} P_{t+s}(x,A) = \int_0^\infty P_t(x,\text{d}y) P_s(y,A) \qquad \forall x \in \mathbb{R}_+ \quad \forall A \in \mathcal{B}(\mathbb{R}_+). \end{equation}
Below what I've done so far.
Observe that $$ P_t(x,A) = e^{-t/x}I_A(x) + \int_x^{\infty} \frac{t}{y^2} e^{-t/y}I_A(y) \, \text{d}y. $$
The LHS of \eqref{eq:ck}: $$ P_{t+s}(x,A) = e^{-(t+s)/x}I_A(x) + \int_x^{\infty} \frac{t+s}{y^2} e^{-(t+s)/y}I_A(y) \, \text{d}y. $$
The RHS of \eqref{eq:ck}: $$ \int_0^\infty P_t(x,\text{d}y) P_s(y,A) = P_tP_s(x,A). $$ Using \eqref{eq:identity}: \begin{multline*} P_tP_s(x,A) = e^{-t/x} \left( e^{-s/x} I_A(x) + \int_x^\infty \frac{s}{z^2} e^{-s/z} I_A(z) \, \text{d}z \right) \\ + \int_x^\infty \frac{t}{y^2} e^{-t/y} \left( e^{-s/y} I_A(y) + \int_0^\infty \frac{s}{z^2} e^{-s/z} I_A(z) \, \text{d}z \right) \, \text{d}y \\ = e^{-(t+s)/x} I_A(x) + e^{-t/x} \int_x^\infty \frac{s}{z^2} e^{-s/z} I_A(z) \, \text{d}z + \int_x^\infty \frac{t}{y^2} e^{-(t+s)/y} I_A(y) \, \text{d}y \\ + \int_x^\infty \frac{t}{y^2} e^{-t/y} \left( \int_0^\infty \frac{s}{z^2} e^{-s/z^2} I_A(x) \, \text{d}z \right) \, \text{d}y \end{multline*}
but then I'm stuck.
Any hint would be appreciated.