Say we have some non-Markovian process, $$X_{t}$$ which depends on some identifiable information contained in the sigma algebra generated by the collection of random variables up to time $$s, $$\mathcal{F}_{s}^{X}$$, such that $$P(X_{t}|\mathcal{F}_{s}^{X}) \not = P(X_{t}|X_{s})$$. Should one be able to extend the state space to support another process, say $$V_{t}$$, such that the all information for which $$X_{t}$$ was dependent on in ($$\mathcal{F}_{s}^{X}- \sigma(X_{s})$$) is encoded in it and such that $$V_{t}$$ is markov, is the process $$(X_{t},V_{t})$$ Markov in the product space? I have seen this alluded to and it makes intuitive sense, though I have not seen this property stated formally.
An example I've seen in the wild would be a M/G/1 queue which does not have an exponential service time but $$(X_{t},V_{t})$$ being the double which records the current queue count ($$X_{t}$$) and the elapsed service time of the current job ($$V_{t}$$) being Markovian.
Yes, you can take $$V_t=(1[s\leq t]\cdot X_s)_{s}$$, so that the state space of $$V_t$$ is now a space of functions (although I will purposely avoid saying what $$\sigma$$-algebra to put on this huge space to avoid getting into the weeds of measurability technicalities...) and the entire process $$(V_t)_t$$ is a path in a function space, or equivalently $$V$$ is a random field indexed by $$V_{s,t}$$.