Why can the acidic strength between H2S and HCl be explained on the basis of electronegativity but H2S and H2O can't?
You are right that electronegativity is an important consideration but another factor that is also very important is size of the atom. Although there is a decrease in electronegativity from $\ce {O}$ to $\ce {S}$, the size of the atom which the charge is mainly centred on is now larger. This allows the charge to be distributed over a larger volume of space, decreasing inter-electronic repulsions, providing greater stabilisation of the negative charge. 

However, the greater size decreases charge density, decreasing the strength of ion-dipole interactions between the anion and the surrounding water molecules. Thus, the extent of solvation decreases for the larger anion. 

Electronegativity is defined as the ability of the atom to attract a bonding pair of electrons. It merely serves as a proxy for the ability of the atom to stabilise negative charge. It should not be viewed as the most important factor that influences the atom's ability to stabilise negative charge. 

Also, mentioned in your answer, the strength of the bond formed with the larger atom with more diffuse orbitals decreases. A bond that is more easily cleaved would facilitate bond cleavage and thus, favour the acid with the larger atom bonded to $\ce {H}$ as the stronger acid. 

Consolidating the above discussion, there are really two factors on each side of the table below. Perhaps, the distribution of charge increases very significantly and the bond energy also decreases significantly while the decrease in extent of solvation and the electronegativity exhibit a relatively less significant effect. It would be great if someone is able to quantitatively show this...  

\begin{array}{|c|c|c|c|} \hline
\text{Favouring acids with larger atom} &amp; \text{Favouring acids with smaller atom} \\ \hline
\ce{Distribution of charge} &amp; \ce{Extent of solvation}\\ \hline
\ce{Bond strength} &amp; \ce{Electronegativity}\\ \hline
\end{array}