UPDATE: This is at best a partial answer.

If $X$ is a Markov process with continuous paths, then the additional condition for it to be a diffusion it be strong Markov (see the book by Ito-McKean, Section 3.1). For example, your example is not strong Markov (stop it when it hits 0). In that case, however, it is still not necessarily Feller (this was stated previously).

If you only require $X$ to be a Markov process with not necessarily continuous paths, then your condition is (almost, see this thread) equivalent to requiring that 1) $S_t:C_0(E) \rightarrow C_0(E)$, where $(S_t)_{t\ge0}$ is the semigroup of $X$ and $C_0(E)$ is the space of continuous functions vanishing at infinity. But a Feller process requires moreover that 2) $S_tf(x) \rightarrow f(x)$ for each $x$, as $t\rightarrow 0$ (see for example Section III.3 in the book by Revuz-Yor). For example, take Brownian motion on $\mathbb R^+$ which jumps to $1$ when it hits $0$. This process satisfies 1) but not 2)

If you take for $X$ any stochastic process, I doubt that one can tell more than what you wrote in your comment.

UPDATE: This is at best a partial answer.

If $X$ is a Markov process with continuous paths, then the additional condition for it to be a diffusion it be strong Markov (see the book by Ito-McKean, Section 3.1). For example, your example is not strong Markov (stop it when it hits 0). In that case, however, it is still not necessarily Feller (this was stated previously).

If you only require $X$ to be a Markov process with not necessarily continuous paths, then your condition is (almost, see this thread) equivalent to requiring that 1) $S_t:C_0(E) \rightarrow C_0(E)$, where $(S_t)_{t\ge0}$ is the semigroup of $X$ and $C_0(E)$ is the space of continuous functions vanishing at infinity. But a Feller process requires moreover that 2) $S_tf(x) \rightarrow f(x)$ for each $x$, as $t\rightarrow 0$ (see for example Section III.3 in the book by Revuz-Yor). For example, take Brownian motion on $\mathbb R^+$ which jumps to $1$ when it hits $0$. This process satisfies 1) but not 2)

If you take for $X$ any stochastic process, I doubt that one can tell more than what you wrote in your comment.

If $X$ is a Markov process with continuous paths, then the additional condition for it to be a diffusion and hence a Feller process is that it be strong Markov (see the book by Ito-McKean, Section 3.1). For example, your example is not strong Markov (stop it when it hits 0).

If you only require $X$ to be a Markov process with not necessarily continuous paths, then your condition is (almost) equivalent to requiring that 1) $S_t:C_0(E) \rightarrow C_0(E)$, where $(S_t)_{t\ge0}$ is the semigroup of $X$ and $C_0(E)$ is the space of continuous functions vanishing at infinity. But a Feller process requires moreover that 2) $S_tf(x) \rightarrow f(x)$ for each $x$, as $t\rightarrow 0$ (see for example Section III.3 in the book by Revuz-Yor). For example, take Brownian motion on $\mathbb R^+$ which jumps to $1$ when it hits $0$. This process satisfies 1) but not 2)

If you take for $X$ any stochastic process, I doubt that one can tell more than what you wrote in your comment.

UPDATE: This is at best a partial answer.

If $X$ is a Markov process with continuous paths, then the additional condition for it to be a diffusion it be strong Markov (see the book by Ito-McKean, Section 3.1). For example, your example is not strong Markov (stop it when it hits 0). In that case, however, it is still not necessarily Feller (this was stated previously).

If you only require $X$ to be a Markov process with not necessarily continuous paths, then your condition is (almost, see this thread) equivalent to requiring that 1) $S_t:C_0(E) \rightarrow C_0(E)$, where $(S_t)_{t\ge0}$ is the semigroup of $X$ and $C_0(E)$ is the space of continuous functions vanishing at infinity. But a Feller process requires moreover that 2) $S_tf(x) \rightarrow f(x)$ for each $x$, as $t\rightarrow 0$ (see for example Section III.3 in the book by Revuz-Yor). For example, take Brownian motion on $\mathbb R^+$ which jumps to $1$ when it hits $0$. This process satisfies 1) but not 2)

If you take for $X$ any stochastic process, I doubt that one can tell more than what you wrote in your comment.