Electrical – How much will the output voltage of a transformer drop as I add loads

External transformer will work well. If you can find a mains to suitable multiple low voltage output windings transformer you can make your own step down. eg a 12/12/6 can be arranged as a 24:6 which is about right.

SMPS is a good idea due to efficiency.
Slightly more dangerous load safety wise - see below.

eg ye olde MC34063

This would cost a few dollars.
Note that with ANY non isolated circuit protection should be provided that will blow a fuse if the converter fails shirt circuit. Running yuour LCD on 24V would be a bad idea. See fig 24 in the above datasheet that shows an isolated output converter.

Resistor dropper + regulator Cheap and nasty solution that works OK

Note that this is quick and easy and nasty. Not recommended but doable.

Rectify 24VAC to about 34 Volts (!)
Pass this through a 47 ohm 20 Watt resistor (can be several smaller resistors in suitable series / parallel arrangement) and then
connect this to a 5V, 1A regulator such as LM340, 7805 etc.

Resistor max drop = 0.5A x 47R =~ 24 V.
V into regulator = 34-24 = 10V max.

Power in resistor = I^2 x R = 0.25 x 48 = 12 Watt.
Use much larger rating eg 20W+

You can tailor R so that V in to regulator is about 8V. Less may cause troubles.
You need a cap at regulator input.

Heatsink regulator appropriately.

AFAIK transformers are disigned to withstand a 180% current overload for short periods. (They can be overloaded this way for something like 15 minutes every 24 hours or so.) But in the long run overloading it with 200% rated current will most likely cause serious overheating and eventually damage. Maybe a good (meaning in your case probably water - or oil - cooling, which can be way more expensive than a new transformer) can provide the necessary cooling to enable the trafo to carry that kind of current.

Additional info: If your supply provides 0.1 A DC as you say, the AC current (RMS value) that comes out of the trafo is not the same as the magnitude of the DC current! In general, assuming the output current of the trafo is not distorted (meaning it keeps its beautiful sine shape - not necessarily true) you have for full-wave diode rectifiers (from: Mohan - Power Electronics):

\$I_{out(DC)}=\frac{2}{\pi}\sqrt{2}I_{sin(AC)}=0.9I_{sin(AC)}\$

meaning that the AC current will be actually 0.22 A if you want 0.2 A DC. But, since the trafo is seriously overloaded, I'm sure the current will be strongly distorted due to the saturation of the trafo iron core. At the end of the day, this will lead to extra heat generation and some other odd behavior.

All in all I don't really think it's a very good idea.

The above calculation applies only if you have a line frequency transformer (50 or 60 Hz) followed by a rectifier. If the circuit includes a high frequency transformer, the situation can be entirely different. So, if this is the case, please edit your question, so we can adjust the answer to that.