Electronic – Parallel resistor in series with LEDs

No. No. No. The current you get with no resistors will be very hard to predict, and will only by the greatest of chances be what you want (and small variations in the 5V and temperature will greatly affect whatever current you happen to get).

Use one resistor per LED and be safe.

You could put them in series and use a single resistor but the supply voltage should be more like like 9V in that case (and the resistor value calculated accordingly).

The power supply is 3v at 60mAh (CC/CV)

A power supply can't force both the current and voltage to specific values at the same time. It can either force a certain voltage and let the load determine the current, or force a certain current and let the load determine the voltage.

Remember, the load has its own I-V characteristic that it must obey. For example, a resistive load obeys Ohm's Law \$V=IR\$.

What a 3-V 60-mA CC/CV supply does is force 3 V, unless doing so would require more than 60 mA, in which case it just provides 60 mA at whatever voltage (lower than 3 V) it takes to make that happen.

Assuming your LEDs' forward voltage is less than 3 V, your supply will operate as a 60-mA current supply, and the scenario will play out much as you have described it.

My scenario: 3v power supply directly from a battery.

A battery is more like a constant-voltage supply (with a series resistance), so this will indeed act differently than the constant-current scenario above.

Shouldn't the other two LEDs still draw 20mAh each?

Yes, they will continue to draw whatever it is they draw at 3 V. If they're the same LEDs as in the constant-current supply scenario (where we hypothesized that the forward voltage is lower than 3 V), then they will draw much more than 20 mA because you're driving them well above their rated forward voltage.

Also, as they heat or cool, their I-V curves change. So one might be okay running on a 3 V battery when first connected, but then start drawing more current as it warms up and eventually (or very quickly) burn itself out.

Boohoo, one LED breaks. That should be all. Why not?

With a constant voltage supply, one LED failing wouldn't affect the others.

With a real battery, you have to remember the internal resistance of the battery. If one LED fails, then that will tend to increase the actual output voltage of the battery after accounting for internal resistance, and that could cause the other LEDs to fail.