Let $(X_t)_{t\ge0}$ be a real-valued Lévy process. Note that $$\mu_t:=\mathcal L(X_t)\;\;\;\text{for }t\ge0$$ is a continuous convolution semigroup$^1$. Let $$\tau_x:\mathbb R\to\mathbb R\;,\;\;\;y\mapsto y+x.$$ $(X_t)_{t\ge0}$ is a time-homogeneous Markov process with transiton semigroup $$\kappa_t(x,B)=\tau_x(\mu_t)(B)=\mu_t(B-x)\;\;\;\text{for }(x,B)\in\mathbb R\times\mathcal B(\mathbb R)\text{ and }t\ge0.$$ If $f:\mathbb R\to\mathbb R$ is bounded and uniformly continuous, it's easy to see that $$\left\|\kappa_tf-f\right\|_\infty\xrightarrow{t\to0+}0\tag1.$$ So, $(\kappa_t)_{t\ge0}$ is a strongly continuous contraction semigroup on the space $U$ of those $f$ equipped with the supremum norm.

Now assume the characteristic function $\varphi_\mu$ of $\mu:=\mu_1$ has the form $\varphi_\mu=e^\psi$, where $$\psi(\xi)=-\frac{\sigma^2}2\xi^2+{\rm i}b\xi+\int e^{{\rm i}\xi }x-1-1_{(-1,\:1)}(x){\rm i}\xi x\:\nu({\rm d}x)\;\;\;\text{for all }\xi\in\mathbb R$$ for some $b,\sigma\in\mathbb R$ and a $\sigma$-finite measure $\nu$ on $\mathbb R$ with $\nu(\{0\})=0$.

Let $$(Lf)(x):=\frac{\sigma^2}2f''(x)+bf'(x)+\int f(x+y)-f(x)-1_{(-1,\:1)}(x)yf'(x)\;\nu({\rm d}y)$$ for $f\in C^2(\mathbb R)\cap\mathcal L^1(\nu)$.

Let $A$ denote the generator of $(\kappa_t)_{t\ge0}$ and $f\in C^2(\mathbb R)$ such that $f,f',f''\in U$. I know several references showing that $f\in\mathcal D(A)$ and $Af=Lf$ either using an appropriate decomposition of $(X_t)_{t\ge0}$ or by considering Fourier transforms.

I would really like to know if we are able to prove the claim by showing that $\left(f(X_t)-\int_0^t(Lf)(X_s)\:{\rm d}s\right)_{t\ge0}$ is a martingale$^2$. Or maybe by a more semigroup-theoretic approach.

$^1$ i.e. $\mu_{s+t}=\mu_s\ast\mu_t$ for all $s,t\ge0$ and $$\int f\:{\rm d}\mu_s\xrightarrow{s\to t}\int f\:{\rm d}\mu_t\;\;\;\text{for all }f\in C_b(\mathbb R)\text{ and }t\ge0.$$

$^2$ Maybe one can use that a process $M:=X-\int_0^{\;\cdot}Y_s\:{\rm d}s$ is a martingale iff $N_t:=e^{-\lambda t}X_t+\int_0^te^{-\lambda s}(\lambda X_s-Y_s)\:{\rm d}s$ is a martingale.