MCU + LED + SSR + (how to determine resistor value)
3.3V - 1.8V (Vf of the led) - 1.15V (Vf of led in SSR) = 0.25V (So my resistor needs to handle 0.25V?)
  
  3mA is needed to activate the SSR, and it looks like 2mA is needed to
  make the led light up, so I am guessing 5mA.


Current is the same for all the elements that are in series, but voltage is dropped at each element. (In parallel the voltage is the same, but the current is shared/divided).

Assuming your LED is rated for 2mA and that your SSR requires 3mA to turn on, you cannot put them in series without:


Exceeding the recommended current of the LED, or 
Not be guaranteed to
turn the SSR on


Both are unwanted scenarios, so you have to go with another route.

Considering that a GPIO on your MCU can source 20mA, even if we stay within the desired ~10mA, you could wire them in parallel:







simulate this circuit &ndash; Schematic created using CircuitLab

Calculating R LED:

Vled = Vtotal - Vforward

3.3V - 1.8V = 1.5V

U = R * i

1.5V = Rled * 2mA

Rled = 1.5V/0.002A = 750R (Which is a standard value, so you can use this or use the next higher one)

Calculating R SSR:

If we aim at 5mA (3mA + 2mA margin), datasheet says about 1.1Vf.

So, same thing:

(3.3V - 1.1V) = Rssr * 5mA

Rssr = 2.2V / 0.005A = 440R (which isn't a standard value, so you can use 420R which should give a bit more than 5mA, or 470R which will give a bit less)



So, in total, you have 2mA for the LED + ~5mA for the SSR, so about 7mA coming from your pin.







simulate this circuit

So, in summary, you got Ohms law right, but you got the series current wrong. :)