The quick answer is that a 20W LED light and a 20W heating element both produce 20W of heat, and inside a black (opaque) balloon, they would both produce the same amount of heat to lift the balloon.
Now, it is obvious this doesn't tell the whole story; intuition seems to suggest a heating element would be more fit for producing heat than a device intended to produce light, not to mention the differences in the weight between the devices.
A lot of people have mentioned how LED lighting isn't as efficient as many people believe.
While it is inaccurate to say LED lighting turns most of its input power into light, it's not like the claim has no basis in reality whatsoever, and I feel like this makes it a bit of a red herring in the reasoning behind the answers.
LED's usually do turn most of its consumed power into light (if more than half counts as most). It's the power circuitry that drives the LED that accounts for most of the power lost as heat.
I have more experience working with laser diodes than LED's, but most often over 60% of the power used in the diode is turned into radiation, while the process of modulating the current flowing through the LED is hugely less efficient.
In the simplest designs, a 20W light might have a power stage that dissipates 18W as heat while supplying 2W to the actual LED, which might produce 1.5W as visible light. Conceivably, you could argue both that the efficiency is 75% (1.5/2) and that it is 7.5% (1.5/20).
(The issues that arise when dealing with thermal dissipation are more attributable to the relatively small size of LED's rather than the amount heat being produced. I realize laser diodes are significantly different than LED's, but I'm assuming most LED's are more efficient, but I may be wrong here)
So in many cases the comparison between a 20W LED light and a 20W (resistive) heating element basically boils down to one between an expensive, unnecessarily convoluted 20W heating element and a 20W heating element, and as people have discussed already, convection would be the more important issue. I would imagine the optimal ohmic heating element in this situation would be a motor of some size with a fan attached to its spindle in a way that it cools the motor.
But, and I think this may be closer to the intent of the question, if you are asking, "assuming you could turn 20W of electricity into either 20W of light or 20W of heat, which would be more effective in producing lift in a hot air balloon?", the answer is, it depends.
You've already taken the color of the balloon into account by saying it's black. This would contain the light produced inside the balloon, but as you've stated, a black balloon would absorb the light and heat up.
To produce lift, we don't want the balloon to heat up. We want the air inside the balloon to heat up.
So let's make the balloon shiny, i.e. 100% reflectivity (this could be approached by using metallized mylar). Then, you keep all the light inside the balloon, and it doesn't waste heat by heating up.
Now that we have all this light energy trapped in the balloon, we just need to turn it into heat in the air. Just kidding, no we don't. The light, not being able to escape the balloon would eventually be absorbed by the molecules in the air itself. It's as it's a law or something, light will eventually turn into heat at some point, in some thing. This heats up the entire volume of the air simultaneously. No convection required. Light clearly wins in terms of speed in this case.
Depending on the level of material sciences, creating a balloon with near perfect reflectivity that is both light and strong enough to support whatever load it carries may be impossible.
Let's go back to black. Probably nylon, or some other fabric. Then, the balloon would heat up, absorbing light better than air, and while it would heat the air inside the balloon, it would also heat the air outside (i.e. lose heat).
But really, this is unavoidable no matter what you do. Even if you used a heating element to heat up the air inside the balloon, once the air is hot, it would heat up the balloon, which would lose heat to the outside.
The main difference would be in how long it takes initially to get the balloon off the ground. It would take some ugly calculus to model (if you heat the air with light via the balloon, you have more surface area = faster heat conduction, but you lose half to the outside and the temperature is lower, and fluid dynamics I don't even want to start thinking about), but put simply, using light would mean a longer wait before the balloon lifted off than using heat, but once the air was at the target temperature, there would be no difference in efficiency.
These thought experiments seemingly pointing to the conclusion that light and heat would be equally as efficient can probably be traced (in a very roundabout way) to one of these facts: light is heat, or eventually becomes heat; where there is heat, there is light; a large part of heat (or all of it, depending on the scale you think at) transmission happens through light; and most fundamentally, heat is such a vague word that it is used in countless contexts and meanings, yet it is difficult to define clearly. Also c.f. black body radiation.
None of this would work with a 20W anything, by the way.