When are stable continuous time Markov chains Feller? Always? This is a question is similar to this 2 year-old one that never got answered  (truthfully it's pretty much the same question except that I'm adding a bit more detail and the assumption that the $Q$ matrix is stable meaning that its diagonal entries are strictly greater than $-\infty$).
I suspect the answer to this question is already in the literature I just haven't stumbled across it yet, if this is the case any such references would be great. Thanks in advance for any replies.
First I add some background (sorry it's a bit long, I'm trying to make sure that we are all talking about the same thing), then I add my questions and finally some of my thoughts.

Background:
Let $E$ be some countable set and $q$ be a "stable $Q$-matrix on $S$":


*

*$q:E\times E\to\mathbb{R}$,

*$-\infty<q(x,x)\leq0$,

*$q(x,y)\geq 0$ for all $x\neq y$,

*$\sum_y q(x,y)=0$ for all $x$.


Consider Kolmogorov backward equation 
$$\frac{d}{dt}p_t(x,y)=\sum_z q(x,z)p_t(z,y)$$
It is well known that there is a unique minimal sub-stochastic transition function $p^*_t:[0,\infty)\times E\times E\to[0,1]$ that solves the above. It is sub-stochastic in the sense that $\sum_y p^*_t(x,y)\leq 1$ and minimal in the sense that any other transition function $p$ that solves the above is such that 
$$p_t(x,y)\geq p_t^*(x,y)$$
for every $t\geq0$ and $x,y$ in $ E$.
I'm interesting in finding out when it is the case that $p_t^*$ defines a "Feller-Dynkin" semigroup (using the terminology of Williams and Rogers) $P_t$. That is, associating the discrete topology to $E$, for any $f\in C(E)$ (where $C(E)$ is the of functions that vanish at infinity)
$(P_t f)(x):=\sum_y p_t(x,y)f(y)$
is such that:


*

*$P_t f\in C(E)$,

*$0\leq f\leq 1$ implies that $0\leq P_tf\leq 1$

*$P_sP_tf=P_{s+t}f$ for all $s,t\geq0$ and $P_0f=f$,

*$||P_tf-f||\to 0$ as $t\downarrow 0$ (where $||\cdot||$ is the supremum norm on $C(E)$).



Questions:
When is $P_t$ a Feller-Dynkin semigroup on $C(E)$? In particular what conditions on $q$ can we impose to ensure this is the case (apart from $\sup_x q(x,x)<\infty$)? The second and third bullet points follow directly from the fact $p_t^*$ is a sub-stochastic transition function, but the first and final ones are (at least to me) not obvious.

Thoughts/attempt:
My suspicion/hope is that the answer is "always" however I've been unable to find this in the literature nor to prove it myself. In particular, I tried applying the Hille-Yosida theorem to $q$ but I got stuck trying to prove that some recurrence equations always had a unique solution (I was also assuming that every row of $q$ had finitely many non-zero entries).  Furthermore I haven't even checked that the semigroup promised by the Hille-Yosida theorem coincides with that defined by $p_t^*$.
 A: It appears that the transition semigroup need not be Feller. An example:
Denote the non-negative integers by $\mathbb{N}$. Define, for $i,j\in\mathbb{N}$, 
$$q(i, j) =\begin{cases} 0 &\mbox{ if }i =0\\
       i^2(\delta_{i-1, j}-\delta_{i,j}) &\mbox{ otherwise.}\end{cases}$$
Then the backward equations are
$$p_t'(0,0) = 0$$
and
$$p_t'(i, i-1) = i^2(p_t(i-1, i-1) - p_t(i, i-1)),$$
for $i$ positive.
Inductively, we see that there is only one solution satisfying, for each $i$, $p_0(i,i)=1$. Let $(X_t)_{t\ge 0}$ be an $\mathbb{N}$-valued process. For $i\in\mathbb{N}$, let $\mathbb{P}_i$ be a law under which $X$ starts from $i$ and has transition semigroup $p$.
Define $H=\inf\{t:X_t=0\}$. Then
$$\mathbb{E}_i[H] = \sum_{j=1}^i j^{-2}.$$
So, there are constants $M>0$ and $\epsilon>0$, such that, for each $i$,
$$\mathbb{P}_i[H\le M] > \epsilon.$$
Define $f\in C(\mathbb{N})$ by
$$f(i) = \begin{cases}1 & \mbox{ if } i = 0\\
                      0 & \mbox{ otherwise.}\end{cases}$$
$$\begin{align}\mathbb{E}_i[f(X_M)] &= \mathbb{P}_i[X_M=0]\\
                       &= \mathbb{P}_i[H\le M]\\
                       &\ge \epsilon.\end{align},$$
thus $P_Mf$ doesn't vanish at infinity, so the first bullet point isn't satisfied.
A: EDIT: This might be a bit late, but it gives you a better answer (than my previous one) I think:
In Ethier and Kurtz (Markov Processes), chapter 8, the Theorem 3.1 or rather its corollary 3.2. gives you the conditions you want. In particular if your $E=\mathbb{N}_0$ and
$$\sup_{i \ge 0}\frac{q_{ii}}{i+1}<\infty,$$
$$\lim_{i \rightarrow \infty}q_{ij}=0, \textrm { for all } j \ge 0,$$
$$\sup_{i\ge 0}\sum_{j \ge 0}\frac{i+1}{j+1}q_{ij}<\infty$$
and
$$\sup_{i \ge 0}\frac{1}{i+1}\sum_{j \ge 0}(j-i)q_{ij}>\infty,$$
then your process is Feller. Of course, if $E$ is not $\mathbb{N}_0$ but still countable, then it is isomorphic to $\mathbb{N}_0$ and so you can easily adapt it to your case or use the more general Theorem 3.1.
You can notice that the process from Ben's counterexample fails to meet the first condition. The second condition does something similar - for example it excludes a process on $\mathbb{N}_0$ for which $q_{i0}=q_{i}=1$ for all $i \neq 0$ which has the same problem as the other counterexample.
Old answer:
Well, it may be too restrictive (and perhaps obvious to you), but one condition that gives you (on top of what you have already assumed) the Feller property is that the generator is bounded (which is not the case in the counterexample given above).
In that case one can just use the Hille-Yosida theorem.
Let's call the generator $A$, i.e. $Af(x) := \sum_{y \in E}q(x,y)f(y)$.
1) $A1\equiv0$, obviously.
2)$\mathcal{D}(A)$ is just $\mathcal C (E)$.
3)$\mathcal{R}(I-\lambda A)=\mathcal C (E)$ for $\lambda$ small enough, since for any $g \in \mathcal C (E)$ one can put
$$f :=\sum_{n=0}^{\infty}\lambda^nA^ng,$$ which converges for $\lambda ||A|| < 1$ (here we need the boundedness of $A$), and which solves the equation $f-\lambda Af=g.$
4) Dissipativity of $A$:
Put $x_0:=\arg \max_{x\in E}|f(x)|$. Then
\begin{align*}
||\lambda f(\cdot)-Af(\cdot)||& = \big|\big|f(\cdot)\big(\lambda+\sum_{y \ne \cdot}q(\cdot,y)\big)-\sum_{y \ne \cdot}q(\cdot,y)f(y)\big|\big|\\
&\ge\big|f(x_0)\big(\lambda+\sum_{y \ne x_0}q(x_0,y)\big)-\sum_{y \ne x_0}q(x_0,y)f(y)\big|\\
&\ge\big|f(x_0)\big(\lambda+\sum_{y \ne x_0}q(x_0,y)\big)-f(x_0)\sum_{y \ne x_0}q(x_0,y)\big|=\lambda||f||.
\end{align*}
In particular, if $E$ is finite then the chain will always be Feller.
