Is it possible to generalize the proof of the statement below to general Lévy kernels $\rho(r)dr$ or even Lévy-type kernels $\rho(x,r)dr$ (in the sense of Lévy Matters III by Böttcher et al.)?
Otherwise, can you give an alternative proof of the general cases (or a generalizable proof)?
The density of the random variable $X^x(\tau_0(x))$ is given by $$ \mathbf P[X^x(\tau_0(x))\in dz]=\left(\int_0^x \rho(y-z) \int_0^\infty p^x_s(y)dsdy\right) dz,\quad \text{on }(-\infty,0), $$ where $X^x(s)$ is the inverted $\beta$-stable subordinator started at $x$, we define $\tau_0(x):=\inf\{s>0:\ X^x(s)\le 0\}$, we denote the density of $X^x(s)$ by $p^x_s$, and $\rho(z)=-\Gamma(-\beta)^{-1}z^{-1-\beta}$.
Proof. Fix $x>0$. Recall that the generator $\mathcal G$ of $X^x(s)$ acts on smooth functions as $$\mathcal Gf(x)=\int_0^\infty (f(x-z)-f(x))\rho(z)dz,\quad \text{and}\quad \mathbf E[\tau_0(x)]=\frac{x^{\beta}}{\Gamma(\beta+1)}.$$ Let $\phi:\mathbb R\to\mathbb R$ be a smooth function compactly supported in $(-\infty,0)$. Apply Dynkin formula and the non-increasingness of the process $X^{x}(s)$ to obtain \begin{align*} \mathbf E[\phi(X^x(\tau_0(x)))]&=\mathbf E\left[\int_0^{\tau_0(x)}\mathcal G\phi(X^x(s))ds\right]\\ &=\int_0^x\mathcal G\phi(y)\left(\int_0^\infty p^x_s(y)ds\right)dy\\ &= \int_0^x \left(\int_y^\infty \phi(y-z)\rho(z)dz\right) \left( \int_0^\infty p^x_s(y)ds\right)dy\\ &=\int_{-\infty}^0 \phi(z) \left(\int_0^x \rho(y-z) \int_0^\infty p^x_s(y)dsdy\right)dz. \end{align*} Now, for a function $\phi \in C_{\infty}((-\infty,0])\cap\{f(0)=0\}$, take a sequence of smooth functions $\{\phi_n\}_n$ compactly supported in $(-\infty,0)$, such that $\phi_n\to \phi$ uniformly on $(-\infty,0]$. If we can apply Dominated Convergence on the rhs above, we obtain the equality for every $\phi\in C_{\infty}((-\infty,0])\cap\{f(0)=0\}$, hence characterizing the measure of $X^x(\tau_0(x))\mathbf 1_{\{X^x(\tau_0(x))<0\}}$. The candidate density on the rhs is a finite measure as \begin{align*} &\int_{-\infty}^0 1 \left(\int_0^x \rho(y-z) \int_0^\infty p^x_s(y)dsdy\right)dz\\ =&\int_{0}^\infty \rho(z) \left(\int_0^x \mathbf 1_{\{y\le z\}} \int_0^\infty p^x_s(y)dsdy\right)dz\\ =&\int_{0}^\infty \rho(z)\mathbf E\left[\int_0^{\tau_0(x) }\mathbf 1_{\{X^x(s)\le z\}}ds\right] dz\\ =&\int_{0}^\infty \rho(z)\left(\mathbf E\left[\int_0^\infty\mathbf 1_{\{0<X^x(s)\le z\}}ds\right]\mathbf1_{\{x> z\}}+\mathbf E\left[\tau_0(x)\right]\mathbf 1_{\{x\le z\}}\right) dz\\ =&\int_{0}^x\rho(z)\mathbf E\left[\int_0^\infty\mathbf 1_{\{\tau_0(x)>s\ge\tau_z(x)\}}ds\right] dz+\mathbf E\left[\tau_0(x)\right]\int_{x}^\infty \rho(z)dz\\ =&\int_{0}^x\rho(z)\mathbf E\left[\tau_0(x)-\tau_0(x-z)\right] dz+\mathbf E\left[\tau_0(x)\right]\int_{x}^\infty \rho(z)dz\\ = &\frac{1}{\Gamma(\beta+1)}\int_{0}^x (x^\beta-(x-z)^\beta )\rho(z)dz+\mathbf E\left[\tau_0(x)\right]\int_{x}^\infty \rho(z)dz<\infty, \end{align*} hence Dominated Convergence applies. It remains to show that $\mathbf P[X^x(\tau_0(x))=0]=0$. This can be done by explicitly computing $\mathbf P[X^x(\tau_0(x))<0]=1.$
It is possible to describe the distribution of hitting times for a larger classs of processes, but in general the integral expression on the right-hand side is going to look more complicated. For instance if $(X_t)_{t \geq 0}$ is a Feller process with characteristics $(b(x),Q(x),N(x,dy))$ (in the sense of Lévy Matters III), then
$$\mathbb{E}^x \left( e^{-\lambda \tau_D} 1_F(X_{\tau_D-}) 1_E(X_{\tau_D}) \right) = \mathbb{E}^x \left( \int_{(0,\tau_D)} 1_F(X_t) e^{-\lambda t} N(X_t,E-X_t) \, dt \right)$$
for any $\lambda>0$ and $x \in \mathbb{R}^d$; here $\tau_D$ is the first hitting time of a set $D \subseteq \mathbb{R}^d$ and $E \subseteq \mathbb{R}^d$ is such that $\text{dist}(D,E)>0$; the identity remains valid for $\lambda=0$ if $\mathbb{E}^x(\tau_D)$ is finite. The key idea is to use Lévy systems which have been introduced by Ikeda & Watanabe.
The above statement can be found in Remark 2.46, Lévy Matters III, and goes back to this work by Ikeda & Watanabe.