I have touch operated light switches operating LED lights throughout my home. The switch manufacturer claims that I must add a device to circuits containing LED bulbs of 5w or less to prevent flickering. They show their device connected between the neutral and hot wires before the driver and after the switch. Some of the devices "popped" - they suggested to replace the device with a 5w 100k ohm resistor bridging the hot and neutral wire before the LED driver and after the switch. Does this make sense? Is this correct? Is this not a short circuit when the switch is on?
First, you are not allowed to use electronic components in mains electrical wiring. You must use "equipment" which has been tested and certified (listed) for use in mains electrical wiring. Equipment may simply be an electronic component in a safe housing built to a certifiable spec and run through the independent testing lab to test for fire and health safety. TUV or UL are testing labs, CE is not.
A resistor is a resistor, as long as it is made for this purpose and the ohm values are within 50-70%. If Moon Rune Importers there can't provide a Listed resistor, get a resistor from another manufacturer.
If that doesn't work for you, change the smart switch to one that does not have the resistor requirement. Electronic ballasts/drivers are not a surprise. CFLs have been on the market for over 20 years, governments have been expressing their intent to deter/ban incandescents for almost that long, to warn manufacturers to stop making things that aren't compatible with electronic drivers. So they have no excuse.
A 100k resistor on 110/120V will draw as follows:
E=IR 120V = I * 100,000ohm I = 0.0012 amps
W=EI W = 120V * 0.0012A W = 0.144 watts
A 100k resistor on 220/230/240V will draw as follows:
E=IR 230V = I * 100,000ohm I = 0.0023 amps
W=EI W = 230V * 0.0023A W = 0.529 watts
Either approach seems reasonable. I'm not sure where you get a 100k resistor that's listed for use with mains electrical, but 0.5 watts is little enough energy that cooling won't be a big factor.
It will be a short, as you suspect, but it won't be a dead short. Current will be limited to 0.0012 amps or 0.0023 amps, depending on your voltage. This will result in watts of power consumed being under 1 watt.
The only thing that surprises me is that such low current flow would suffice.
The resistor you talk about provides an alternate path for a small amount of current -- these "smart" touch switches apparently use the short-cut of trickling power through a turned-off load in order to power themselves. (Apparently the smart switches only need a couple mA of current, hence the large resistor value used.)
A capacitor is a better solution
However, because of the power dissipation required by a resistor, the more typical solution to the "trickle current" problem with LEDs is to use a load capacitor to provide an alternate path for the small amount of current required by a smart switch. In fact, Lutron makes just such a capacitor as the LUT-MLC, and it's UL listed for use on systems running anywhere between 120 and 277VAC, making it usable basically the world around.
If you really want to use a resistor...
If you really want to use a resistor for this, the resistor will need sufficient wattage (1 or 1.5W) and voltage (500V) ratings, and also must be mounted clear of combustible materials (in the US, NEC 470.3 specifies a 12" clearance from combustibles) as it will get toasty-warm in operation. Temperature rises of 70 or more degrees C can be expected for the resistor in question despite the low current draw.