When are stable continuous time Markov chains Feller? Always?

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.

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.