Neutralization vs Solvation vs Dilution:
Semantically, the reaction of acid or base with water is (usually) solvation and dilution, not neutralization. Pure water $\ce{pH}$ (at $\pu{25^{\circ}C}$) is always 7, i.e. neutral.
When you add an acid to water, you are dissolving the acid. The formation of $\ce{H3O+}$ ions necessarily lowers the $\ce{pH}$, away from neutral:
$\ce{pH \equiv -log10([H+]) = -log10([H3O+])}$
(Technically, $\ce{pH}$ is defined with activity of $\ce{H+}$, but concentration is a good enough approximation for most cases.)
As you can see, higher concentration of $\ce{H3O+}$ ions leads to lower $\ce{pH}$.
Similarly, when you add a base to water, you are increasing the concentration of $\ce{OH-}$ ions (which also destroy some of the naturally occurring $\ce{H3O+}$ ions, lowering their concentration), so $\ce{pH}$ rises away from neutral.
$\ce{pH \approx 14 - pOH = 14 + log10([OH-])}$
Higher $\ce{OH-}$ concentration leads to higher $\ce{pH}$.
When you then add more water, you are reducing the concentration of both $\ce{H3O+}$ and $\ce{OH-}$ ions (as well as that of all other dissolved species), thus bringing the $\ce{pH}$ closer to neutral. This is called dilution.
Neutralization is when you add a base to an already acidic solution ($\ce{pH < 7}$), or an acid to an already alkaline solution ($\ce{pH > 7}$) (one can also react acids and bases out of solution, but that's beyond the scope of this post). In the former case, new $\ce{OH-}$ ions react with pre-existing $\ce{H3O+}$ ions; in the latter case, new $\ce{H3O+}$ ions react with pre-existing $\ce{OH-}$ ions. Either way, the following neutralization reaction occurs:
$\ce{H3O+ + OH- -> 2H2O}$
Thus, the concentration of $\ce{H3O+}$ and $\ce{OH-}$ ions is reduced, and the $\ce{pH}$ is again brought closer towards neutral.
Note that when you dilute a solution with water, the above reaction doesn't occur.
Ions, Salts, and Hydrates:
A salt, by definition, is a chemical compound consisting of an ionic assembly of cations and anions. In solution, these ions dissociate from each other, and are surrounded by solvent molecules (solvation shells). So in solution, what you have isn't exactly a salt, but rather a collection of isolated ions. Some of these can and do form solid salts. This is determined by the salt's solubility product.
The $\ce{H3O+}$ ion will generally not form solid salts, because its affinity to water is too high. It's very similar to plain water molecules. However, if we look at solid samples of various hydrophilic substances, we'll often find that they're hygroscopic. Meaning, they attract water molecules even from air, and form hydrates. This is when a substance's molecules are intermixed with water molecules even in the solid phase.
If we look for example at a solid acid such as benzenesulfonic acid, we will find that it readily forms a hydrate. For very strong solid acids, some of the water molecules in the hydrate will in fact be in the $\ce{H3O+}$ form. But for weaker acids, most of the water molecules will be in the neutral $\ce{H2O}$ form. More info here.
If you add too much water to such acids, the hydrate will dissolve, and you'll get an acidic solution.
On the other hand, $\ce{OH-}$ ions form stable salts much more readily. For example, if you add a base to a solution of $\ce{CaCl2}$, the $\ce{Ca(OH)2}$ will precipitate:
$\ce{CaCl2(s) + H2O -> Ca^2+(aq) + 2Cl-(aq)}$
$\ce{CaCl2(aq) + 2NaOH -> 2Na+(aq) + 2Cl-(aq) + Ca(OH2)(s) v}$
The $\ce{Ca(OH)2}$ is itself a base, but poorly soluble in water. And $\ce{NaOH}$, a common base, is also technically a salt.
Regarding your two examples:
As Poutnik noted, $\ce{HCN}$ is a weak acid and is a gas at very close to room temperature. But if you could isolate your suggested "$\ce{H3OCN}$", you would find this is just the normal hydrate: $\ce{HCN.H2O}$. The water would be in the neutral form. Even in solution, the $\ce{H3O+}$ and $\ce{CN-}$ ions would only appear in trace amounts (since this is a weak acid).
For your 2nd example, $\ce{NH4Cl}$ does indeed exist, but it is a very soluble salt. So if you react say aqueous ammonia with $\ce{HCl}$, you will only get the separate ions:
$\ce{NH3(aq) + HCl(aq) <=> NH4+(aq) + Cl-(aq)}$
You would need to remove water for the salt to precipitate (very carefully, so that it doesn't decompose back to $\ce{NH3}$ and $\ce{HCl}$, which are both gases).
