I'm looking for a more general Feynman-Kac formula that works in the case of jump-diffusion processes. 
I know that, given a pure diffusion process like
$$dS\_t=\mu\_tdt+\sigma\_tdW\_t,$$ if $u(t,s)$ satisfies the PDE
$$f\_t(t,s)+\mu\_tf\_s(t,s)+\frac{\sigma\_t^2}{2}f\_{ss}(t,s)-V(s)f(t,s)=0$$ with terminal condition $f(T,s)=H(S\_T)$
then $u$ is of the form $$u(t,s)=\mathbb{E}\big[e^{-\int\_t^T{V(S\_x)dx}}H(S\_T)|S\_t=s\big].$$
Is it possible to extend such a result when the process dynamics are given by 
$$dS\_t=\mu\_tdt+\sigma\_tdW\_t+\gamma\_tdN\_t$$ with $N$ a Poisson process independent fom $W$?



Feynman-Kac for jump-diffusion
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


