Would it take 7 to 10 times more energy to lift a turbofan-helicopter? I want to figure out how much energy it would take to lift a turbofan-helicopter.

For reference I started with a rotor helicopter and assumed $T=100\ \mathrm{kN}$ gross weight and $d = 16\ \mathrm m$ rotor diameter (compare S92 or AH64).

Using the propeller-formula
$$
T = ( 0.5*P^2*\eta_\text{tot}^2*\pi*d^2*\rho_\text{air})^{1/3}.
$$

with $\eta_\text{tot}^2=0.58$ (source) I get an engine-output of $P=1\,870\ \mathrm{kW}$ 

If I now take a typical turbofan and a take-off $tsfc$ of $10\ \mathrm{g/(kNs)}$ I get a fuel consumption of $1\ \mathrm{kg/s}$ for my $100\ \mathrm{kN}$ helicopter. If I furthermore assume 30% overall efficiency and $E_\text{Jetfuel}=43.5\ \mathrm{MJ/kg}$ I get an energy demand of $13\,050\ \mathrm{kW}$ to run the fan (neglected all the other stuff taking energy from the shaft).

Does this mean it would take 7 times the amount of energy to lift a turbofan-helicopter (with engines running at full throttle)? Or a magnitude if I assume cruise tsfc...

Does this suit with your experience? Or is there a big error/misconception in my short calculation?
Helicopter rotors are very large precisely for that reason. Lift from a rotor (or thrust from a propeller or fan) is obtained by accelerating a given mass of air m to a velocity v. The thrust obtained is proportional to the increase of momentum ∆mv. Obviously, you can get the same ∆mv by accelerating a given mass of gas to a high speed (propeller/fan) or by accelerating a larger mass to a lower speed (rotor). But the energy needed is proportional to v squared... Hence, and for the same thrust/lift, it's much more expensive, in energy or power terms, to use a propeller/fan than a big rotor ...