Is there an electrolyte that will halt the production of Hydrogen in water electrolysis?
The reduction of hydrogen is what the reduction potential of every other reduction potential is based off of, so:

$$\ce{2 H+_{(aq)} + 2e^{–} -&gt;H2_{(g)}}~~~~~\Delta E^{\mathrm{o}}=0\,\mathrm{V}$$

So if a species has a has a reduction potential $&gt;0\,\mathrm{V}$, then it will be be reduced before the naturally occurring $\ce{H+}$ ions are (these exist in solution due to the auto-protolysis of water, $\ce{H2O -&gt; H+ + OH-}$, thereby not producing any $\ce{H2_{(g)}}$.

If your goal is to produce $\ce{O2_{(g)}}$ from the oxidation of $\ce{OH^{-}_{(aq)}}$, the reaction for this is:

$$\ce{4OH^{−}_{(aq)} -&gt; O2_{(g)} + 2 H2O_{(l)} + 4 e^{−}}~~~~~\Delta E^{\mathrm{o}}=-0.401\,\mathrm{V}$$

Similarly, if a species has a has a oxidation potential $&gt;-.401\,\mathrm{V}$, then it will be be oxidized before the  $\ce{OH-}$ ions, thereby not producing any $\ce{O2_{(g)}}$.

Wikipedia has a rather comprehensive standard reduction potential table, so use that to compare the reduction potential of ions. Choose your salts wisely!

To answer your other question, if this reaction continued indefinitely, a very acidic solution would result, as for every $\ce{OH-}$ that is oxidized, a $\ce{H+}$ is left in solution. Once all of the other cation from the salt is reduced, however, the $\ce{H+}$ would then start being reduced, producing $\ce{H2_{(g)}}$.