Is NaCl in water considered "solution" despite of NaCl's having some precipitation?
When we are considering $\ce{HCl(aq)}$ and $\ce{NaOH(aq)}$. the system contain sufficient amount amount of solvent water molecules  in situ already present and this is more than sufficient to hydrate the produced $\ce{NaCl}$ in the reaction.
Hence the first equation
$$\ce{HCl(aq) + NaOH(aq) -&gt; NaCl(aq) + H2O(l)}$$

will be the correct expression.