Presumably because not all jump processes are not Markov jump processes.
Also jump processes do not have discrete space. Take a compound Poisson process, for example, that is a process for which jumps happen at a fixed rate $\lambda$, but the jump distribution is not a constant 1, but instead can be a distribution (which may be continuous), therefore the space is not discrete.
Also there are also jump processes, which has independent increment property (and markov property) but are not compound Poisson process. These processes, with Brownian motions are the only processes with independent increment properties. they are called Levy processes.
Notice jump continuous-time discrete space and compound Poisson process has finite variation, but this is not necessarily the case with general Levy processes (even without the Brownian motion component), which are pure jump processes.
Of course, this is only a small class of Markovian jump processes. I am sure there are plenty of others which I have not mentioned here. Also there are markov processes for which the transition probability is time dependent, but these won't have the Q matrices as generator which you might have had in mind.
Ahhh... A silent downvote, nearly four years after this was posted. So cute...
According to the comments, there is a problem of comprehension at the heart of this question, which might be the following.
Consider a real valued random variable $Z$. This is a measurable function $Z:\Omega\to\mathbb R$ between a probability space $(\Omega,\mathcal F,P)$ and the measurable space $(\mathbb R,\mathcal B(\mathbb R))$. The distribution of $Z$ (aka the law of $Z$) is definitely not the probability measure $P$, but the image of $P$ by $Z$, often denoted $P_Z$, that is, the probability measure $\mu$ on $\mathcal B(\mathbb R)$ defined by the identity $\mu(B)=P(Z^{-1}(B))$ for every $B$ in $\mathcal B(\mathbb R)$.
Thus, to declare that two real valued random variables $X$ and $Y$ are identically distributed means that $X:\Omega\to\mathbb R$ and $Y:\Phi\to\mathbb R$ for some probability spaces $(\Omega,\mathcal F,P)$ and $(\Phi,\mathcal G,Q)$, and that the probability measures $P_X$ and $Q_Y$ on $\mathcal B(\mathbb R)$ coincide. The fact that $P=Q$ or that $P\ne Q$ (both can happen) is quite unrelated.
To declare that two real valued random variables $X$ and $Y$ are such that $X:\Omega\to\mathbb R$ and $Y:\Omega\to\mathbb R$ for some common probability space $(\Omega,\mathcal F,P)$, as you seem to be intent on saying, one usually simply mentions that $X$ and $Y$ are defined on the same probability space.
Best Answer
Basically, a stochastic process is to a time series what a random variable is to a number.
The realization (the "result", the observed value) of a random variable (say, a dice roll) is a number - (but, as it's a random variable, we know that the number can take values from a given set according to some probability law).
The same applies to stochastic process, but now the realization instead of being a single number is a sequence (if the process is discrete) or a function (if it's continuous). Basically, a time series.