I was probably a bit confused because of the operators used. The operator shown in the question is the Kolmogorov forward operator. The backwards operator is the adjoint of $L$, given by
$$
L^* f = \sum_i A_i(x,t) \frac{\partial f}{\partial x_i} + \sum_{i,j}\frac{B_{i,j}(x,t)}{2}\frac{\partial^2 f}{\partial x_i \partial x_j} - c(x,t) f.
$$
Furthermore, the operator only helps to define the process $X_t$, which will lead to the Ito formula and then to the Feynman-Kac formula.
Hi it is possible to get some Feynman-Kac formula in this case. The proof only use the martingale property and Itô's formula for jump-diffusion processes.
So let's have $X$ s.t. (I took the compensated version of your sde):
$dX_t=[\mu(t,X_t)+\lambda(t)\gamma(t,X_t)]dt + \sigma(t,X_t)dW_t+ \gamma(t,X_{t-})d\tilde{N}_t$ where $\tilde{N}_t$ is a compensated Poisson process of intensity $\lambda(t)$.
Please notice the explicit dependence in $t$ and $X_t$ of the above equation that is necessary to have Markov property for the solution which is necessary for the Feynman-Kac theorem to apply.
Now let's us be given $F(t,X_t)=e^{-\int_s^t V(X_r)dr}u(t,X_t)=e^{-IV(s,t)}.u(t,X_t)$ and apply Itô to this formula. You get :
$$dY_t=dF(t,X_t)=e^{-IV(s,t)}\Big(\big(\partial_t u(t,X_t)+\lambda(t)[u(t,X_t+\gamma(t,X_t))-u(t,X_t)]+\mu(t,X_t)\partial_x u(t,X_t)+\frac{\sigma^2(t,X_t)\partial_{xx}u(t,X_t)}{2}-V(t,X_t).u(t,X_t)\big)dt+ (u(t,X_t+\gamma(t,X_t))-u(t,X_t))d\tilde{N}_t+(\sigma(t,X_t)\partial_{x}u(t,X_t))dW_t\Big)$$
Now if the $dt$ term is null then $Y_t$ is martingale and for $t=T$ :
$$Y_s=F(s,X_s=x)=E[Y_T|X_s=x]=E[e^{-\int_s^T V(X_r)dr}H(X_T)|X_s=x]$$
So in this case the PIDE that solves the Feynman-Kac formula is :
$$\partial_t u(t,X_t)+\mu(t,x)\partial_x u(t,x)+\frac{\sigma^2(t,x)}{2}\partial_{xx}u(t,x)+\lambda(t)[u(t,x+\gamma(t,x))-u(t,x)]=V(t,x)u(t,x) $$
With final condition $u(T,x)=H(x)$
Best regards
Best Answer
The idea is to choose a stochastic process of the form $$dX_t=\mu(t,X_t)dt+\sigma(t,X_t)dW_t $$ and consider the process $Y_t=F(t,X_t)$. Applying Ito to $Y_t$ gives $$ dY_t=\left(F_t+\mu(t,X_t)F_x(t,X_t)+\frac12\sigma(t,X_t)^2F_{xx}(t,X_t)\right)dt+\sigma(t,X_t)F_x(t,X_t)dWt $$
And then to choose the coefficients $\mu$ and $\sigma$ in such a way to have $Y_t$ martingale. In our case if we choose $\mu(t,x)=0$ and $\sigma(t,x)=\sigma x$ we will have $$dY_t=\sigma(t,X_t)F_x(t,X_t)dWt$$ under some additional hypothesis of polynomial growth of $F$ we can conclude that $Y_t$ is a martingale and then that $Y_t = E_t[Y_T]$ which we can rewrite $$F(t,x)=E[F(T,X_T)|X_t=x)$$ now solving the SDE $dX_t=\sigma X_t dW_t$ gives $$X_T=X_t \exp(\sigma(W_T-W_t)-\frac12 \sigma^2(T-t))$$
and hence in law we have $$X_T|X_t \sim x \exp(\sigma\sqrt{T-T}Z-\frac12 \sigma^2(T-t))$$ where $Z$ is a standard Gaussian, hence $$F(t,x)= E\left[ \left(\ln(x)+\sigma\sqrt{T-T}Z-\frac12 \sigma^2(T-t)\right)^4\right] $$
this amounts of calculating the four first moments of a standard Gaussian.
Hope this helps