Could it be that the amps flowing through the regulator is too damn high (whatever is connected to out rst and other 2 jumpers)
The limit to asm1117 is I think (read datasheet before trusting me) is 800mA and 78l05 won't help either with it's 100mA limit. 7805 has 1amp limit. But do also know that the limit is not for continuous load.
But do also know that tantalums are sometimes marked with darker line on positive. (blew up 2 of mine before I figured it out :p)
Ceramic capacitors have very low ESR. Tantalum capacitors are somewhere between ceramic capacitors and electrolytic capacitors. Almost all solid (polymer) capacitors have ESR lower than tantalum capacitors.
So the 100nF ceramic capacitor will not help the stability of 1117 regulator, but can help filter at some frequencies.
As I said in my post and as others have said, you either need to
1. use a capacitor series with higher ESR (typically ESR scales with capacitance and volume, hence why I suggested sticking within 47...150uF as less than 47uF could have more than 1 ohm ESR and more than 100-150 can be too low) OR
2. artificially increase the ESR by adding a resistor in series with the capacitor on the output ... if I'm forced to use 1117 regulators with ceramic capacitors I'd use a 0.2 ..1 ohm resistor
But in the end as I suggested in another post above, if you don't need that much current, you can use smarter / better linear regulators which can work with ceramic capacitors which are much cheaper than an electrolytic capacitor or tantalum capacitor and use less space and you can do lower height designs (you may have to lay electrolytic capacitors flat or use multiple lower height one in some designs) and can be better for some circuits (ex less chance of being ripped off the board or leak electrolyte damaging the circuit)
The 1117 are super cheap, for example let's say 10 cents each... and the other ldos may be 25-50 cents each, but include the price of capacitor and the space on board saved and you're nearly at the same price.
Another tip : I could be a good idea to add a resistor in front of the linear regulator, or at least a resistor footprint - you can add a 0 ohm resistor or a jumper link if you don't want to implement this.
The resistor can drop some voltage across it, giving the linear regulator less voltage to work with and therefore heat less.
For example, the schematic has 12v in and 5v out and let's say you want at least 8v to reach the regulator at 0.20 A of current ... You have the formula V = I x R so R = (12v - 8v ) / 0.20 = 4/0.20 = 20 ohm .... so let's say you go with either 18 or 22 ohm resistor
The power dissipated in the resistor will be P = I x I x R = 0.20 x 0.20 x 18 = 0.72 watts so you'd would probably be fine with a 1w rated resistor.
There's 1w rated surface mount resistors in 1206 footprint.. and bigger footprint ones are super cheap :
https://www.digikey.com/short/4fwvv30fAnd now, your regulator only needs to drop 8v-5v = 3v at 0.20a so it would only dissipate 0.6 watts.