I assure you that your worries are without basis. For the hydrolysis of water, the half reactions are:
$$\ce{2H2O_{(l)} ->O2_{(g)} +4H+_{(aq)} +4e-}~~~~\varepsilon^0=-1.23~\mathrm{V}$$
$$\ce{4H+_{(aq)} +4e- ->2H2_{(g)}}~~~~\varepsilon^0=0~\mathrm{V}~\mathrm{(by~definitoin)}$$
The $\varepsilon^0$ of the reaction would be:
$$\ce{2H2O_{(aq)} ->2H2_{(g)} +O2_{(g)}}~~~~\varepsilon^0=-1.23~\mathrm{V}$$
With the proposed half reactions your are suggesting might take place:
$$\ce{2H2O_{(l)} +2e- ->H2_{(g)} +2OH-_{(aq)}}~~~~\varepsilon^0=-0.83~\mathrm{V}$$
$$\ce{2Cl-_{(aq)} ->Cl2_{(g)} + 2e-}~~~~\varepsilon^0=-1.36~\mathrm{V}$$
The $\varepsilon^0$ of the reaction would be:
$$\ce{2H2O_{(l)} + 2Cl-_{(aq)} -> Cl2_{(g)} + H2_{(g)} + 2OH-_{(aq)}}~~~~\varepsilon^0=-2.19~\mathrm{V}$$
(Note that the $\ce{Na+}$ ions are present in the solution, though it is not listed in the reaction)
Because the hydrolysis of water requires less energy than the production of $\ce{Cl2}$, $\ce{H2}$, and $\ce{NaOH}$ from the ions present in the solution, the second reaction will not occur.
The reaction $\ce{H2O_{(l)} + Cl^{-}_{(l)} -> HCl_{(g)} + OH-_{(aq)}}$ would not occur. $\ce{HCl}$ could not evolve from the solution, as there would be no mechanism forcing the $\ce{H+}$ ions onto the electronically stable $\ce{Cl-}$ ions. Rather, it would be much more likely that the $\ce{H+}$ ions would re-react with the $\ce{OH-}$ ions to reform water, meaning no net reaction.
In short, go right ahead using $\ce{NaCl}$ in your hydrolysis reactions. It's actually probably one of the safest salts you could use!