Can we dissolve salts with higher lattice enthalpy by providing heat? It is well-known that

A salt with a higher lattice energy will not dissolve at all, because then $\Delta_{\text{sol}}H^\ominus$ will be negative.

So, the reaction should be endothermic, right? My question is:
If then we provide heat to the reaction, then we should be able to dissolve such salts. Is this true?
Heat itself it just indirect factor. It is temperature what determines the shift of solubility products. If the dissolution is endothermic process, the salt solubility product would increase with higher temperature.
But if the solubility product is still small even at high temperature, like e.g. $\pu{e-16}$ instead of $\pu{e-20}$, then only a tiny extra salt amount would dissolve when heated.
The effective result depends on the particular case. The typical cases where significant solubility increase is observed are  potassium salts  $\ce{KNO3, KClO3, KClO4}$. But they have limited solubility even at low temperature - see the solubility table.
Generally, the dependence of an equlibrium constant of a reaction, including dissolving/precipitation process, is described by van't Hoff equation

$$\frac{\mathrm{d}}{\mathrm{d}T} \ln K_\mathrm{eq} = \frac{\Delta H^\ominus}{RT^2}$$

For solubility products, being equlibrium constants, is valid the following reaction:
$$\Delta G^{\circ}_\mathrm{dis} = \Delta H^{\circ}_\mathrm{dis}  - T \Delta S^{\circ}_\mathrm{dis} = RT \ln K_\mathrm{sp}$$