It would need a 3.6 Ohm resistor to get anywhere near 500mA, from 5.0V supply which would be 1.6W, (assuming negligible increase in Vf), rising to 1.8W at the max USB Vbus of 5.25V. As 3.6 Ohm resistors don't cost any more than 6.8 Ohm resistors, the way to bet is that is is entirely fake and either uses a 1W LED or rejects from a 2W LED line.
It would be possible to characterise the LED's Vf change with temperature, by heating it in an environment chamber, pulsing the current with a very low duty cycle (e.g 1/1000) to minimise self-heating, which would let you estimate the die temperature from the measured Vf for a steady state current the same as the pulsed current, then you could estimate if doubling the dissipation would damage the junction, but the odds are that it isn't worth the effort.
If you post a good macro photo of the LED (+ PCB) on a graph paper background, so we've got the scale, someone with more experience with current LEDs than myself may recognise the package and be able to guesstimate the power from the die area.