 Solution

Starting from $Y_{n-2}=0$ forces $X_{n-2}=0$. The first nonzero value is consequently $X_{n-1}=-1$. Remaining nonzero at the next step forces $X_n=1$, because $p_{-1,-1}=0$. Hence, on an event of positive probability,The two values are unequal by assumption. They expose memory of the hidden sign: immediately after leaving zero the sign is negative, whereas after two consecutive nonzero observations it is positive. Whenever the relevant observed histories coexist with positive probability, the Markov property for $Y$ fails, since its next-step law is not determined by its present value alone. In particular, the absolute-value projection does not preserve the Markov property for all initial laws; the strong lumpability condition would require $p_{-1,0}=p_{1,0}$.

There is a small qualification to a blanket conclusion about an unspecified initial law. The printed assumptions do not ensure that either conditioning event is possible, and a degenerate initial law may never reach the hidden states that cause the failure. For example, in the order $-1,0,1$, takeAll printed constraints hold, yet $Y_n=0$ always and is a Markov chain. Thus the correct general conclusion is the transformed process is not necessarily Markov; the two computed probabilities give the intended failure when their histories are available.