We have
\begin{aligned} E(Z_1) = A \\ \Pr \{ Z_2 = Z_1 + 1 \} = \frac 1 2 \\ \Pr \{ Z_2 = Z_1 - 1 \} = \frac 1 2 \end{aligned}
Are these conditions enough to get $E(Z_2)$?
We have
\begin{aligned} E(Z_1) = A \\ \Pr \{ Z_2 = Z_1 + 1 \} = \frac 1 2 \\ \Pr \{ Z_2 = Z_1 - 1 \} = \frac 1 2 \end{aligned}
Are these conditions enough to get $E(Z_2)$?
Yes, because $\newcommand{\E}{\mathbf{E}}$ $ \E[Z_2] = \E[Z_1] + \E[Z_2-Z_1] $ holds for all random variables $Z_1$ and $Z_2$ (and you can calculate each term on the right). In other words, linearity of expectations can be applied in all cases, however correlated the random variables may be.