You can directly deal with inhomogeneous Markov processes through the Kolmogorov backward and forward equations. I suppose you are asking for the forward equation (i.e. derivative with respect to the time in the future). Let me discuss things on a formal level, which means here I ignore regularity conditions to ensure the existence of generators, derivatives, etc etc.
Let me first give the backward equation, which is probably easier (a specific case of Feynman-Kac formula). Let $u(s,x,t)=\mathbb{E}^{s,x} f(X_t)$ for a nice function $f$ (e.g. infinitely differentiable with compact support), where the expectation is under the measure $\mathbb{P}^{s,x}$ such that $\mathbb{P}^{s,x}\{X_s=x\}=1$. Let $A_s$ be the generator (I am ignoring the detail of how $A_s$ is defined, you can probably figure it out yourself or see references I give below). Then the backward equation states that
$\frac{\partial u}{\partial s}+A_s u =0$.
The forward equation is usually formulated for the density $p(s,x,t,y)$ of the transition kernel $P_{s,t}(x,B)=\int_B p(s,x,t,y)dy$:
$\frac{\partial p}{\partial t}=A_t^* p(s,x,t,y)$
where $A_t^*$ is the adjoint of the operator $A_t$ (again, there's subtlety in how you definite $A_t$ for forward and backward equations separately, I ignore that here).
For a given diffusion given as an SDE $dX_t=\mu(t,X_t)dt+\sigma(t,X_t)dW_t$ for a Wiener process $W$, it is pretty straightforward that $A_t f(x)=\mu(t,x)f'(x)+\frac{1}{2}\sigma^2(t,x)f''(x)$.
All these are discussed to certain extent in "The Theory of Stochastic Processes, Vol II" by Gikhman and Skorokhod, and "Multidimensional Diffusion Processes" by Stroock and Varadhan (expressed in an integral form instead of derivatives).
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
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.