Can the fact that dark energy increases with volume be explained by classical thermodynamics? Considering adiabatic process in classical thermodynamics, a normal substance with (positive) pressure must do work on its environment in order for the volume to increase by $ dV $(like pushing the walls of a piston) so it's internal energy decreases. 
In contrast, in case of an unusual substance with negative pressure, we must do work on it in order for the volume to increase by $ dV $ so it's internal energy increases. 
But in the case of dark energy, the volume of space expands due to the repulsive gravitational effect of negative pressure in general relativity without any work applied to it. 
I’ve read several sources claiming that the reason dark energy increases with volume can be explained by classical thermodynamics. But I don’t see how it is so since the volume of space  increases due to gravitational effects in general relativity, and not from work being done to it like in classical thermodynamics.
Isn’t it illegal to use classical thermodynamics to conclude that the energy of dark energy increases with volume? 
 A: If you want to take the boring way out, you can say that the increase in total dark energy is a consequence of the equation $\nabla_\mu T^{\mu\nu} = 0$, as you can see with about a page of calculation. But not only is that not necessary, it's morally wrong -- one should always try to explain results using the simplest formalism possible. And here there's nothing wrong with the "thermodynamic" argument.

But in the case of dark energy, the volume of space expands due to the repulsive gravitational effect of negative pressure in general relativity without any work applied to it.

The key issue is that how work is "really" done never matters in thermodynamics.
Consider a large box of gas, which is adiabatically expanded by a piston at the top. If the box is large enough, there's no way for a small parcel of gas at the bottom to know there's a piston; all it knows is that its neighbors are getting further apart, allowing it room to expand. To compute the parcel's change in temperature, you just apply $dW = P dV$ to the parcel alone, even though this volume change is not directly caused by a piston. Instead it's mediated by the repulsive intermolecular forces between parcels.
This exact same reasoning holds for a fixed comoving volume of space containing dark energy. All it knows is that the adjacent volumes are giving it room to expand. It doesn't matter why that's happening. There could be a "piston at the end of the universe", or the adjacent volumes could just be receding due to spacetime expansion.
If you're not convinced, consider the filling of a tank familiar to engineers. Suppose you have an tank with vacuum inside. A valve is opened, and air rushes in. The air inside the tank will be hotter than the surrounding air because it's forcibly pushed in by the air pressure. To compute how much hotter, you can imagine the room containing the tank is fitted with an imaginary piston; as air is sucked into the tank the piston can retract, doing $P dV$ work. In reality, air just rushes in the room from outside, and there's no piston anywhere harvesting the work, but the tank does not know or care.

Isn’t it illegal to use classical thermodynamics to conclude that the energy of dark energy increases with volume?

The entire point of thermodynamics is that its validity never depends on the microscopic details. It doesn't care what's "really" going on, it just relates macroscopic measurable parameters to others.
For example, the thermodynamics developed for classical gases in the 1800's works just fine for quantum gases. You just plug in the equation of state and everything holds just as before. So it's not surprising that thermodynamics holds when GR is involved as well.
The only other quibble you might have is that the First Law really says
$$dU = dW + dQ$$
and we haven't accounted for the heat. But there clearly is no heat transfer going on in this situation. More generally, this tells us the expansion of the universe is thermodynamically identical, when considering small comoving volumes, to ordinary adiabatic expansion. For example, since a photon gas has entropy $S \propto VT^3$, we have $T \propto 1/a$ for radiation. This is the easiest way I know of to derive that result.
A: 
Isn’t it illegal to use classical thermodynamics to conclude that the energy of dark energy increases with volume?

According to thermodynamics the internal energy $U$ doesn’t change in case of adiabatic free expansion (means outside the piston is vacuum) of an ideal gas, so $dU = o$. What changes is density of the gas, it decreases.
This is much different in general relativity if we consider the accelerated expansion of the universe driven by the cosmological constant. In this case the vacuum density $\rho_v$ remains constant if the volume of the piston increases and hence the internal energy increases which yields $dU = \rho{_v}dV$. Now following thermodynamics the work done by the vacuum pressure is $W = -p{_v}dV$. That clarifies that the vacuum pressure is negative. One could say it is taken from an infite reservoir.
So nothing is illigal, but one has to accept that the energy density of the cosmological constant doesn’t get thinned like matter during expansion.
