How do I put an LED in series with an optotriac?
You can put a LED in series with the LED in the solid state relay.  That will make the total voltage you need to drive the combination higher, but as long as you have the voltage available that's no problem.

Let's say you want to use a green LED to indicate the SSR is being driven on, and you have a 5V output to turn on the SSR.  Green LEDs drop about 2.1V.  You should be easily able to find one rated for 20mA, as most are.  Probably the 20mA spec is a maximum for the SSR (for you to check), so let's say we want 10mA going thru the string.  The LEDs will drop 1.3V + 2.1V = 3.4V, leaving 1.6V for the resistor.  1.6V / 10mA = 160&Omega;

So, put the 160&Omega; resistor, LED, and SSR input all in series.  Now you can drive the whole thing with 5V to turn on the SSR and light the indicator, and 0V or open for off.  When on, it will draw about 10mA.

Added:

In the description above the same current is flowing thru the indicator LED as the SSR LED.  It looks like this SSR wants fairly large drive current of around 20mA.  While you can certainly find a LED that can handle that, most "normal" and cheap LEDs are rated for 20mA max.  They will also be plenty bright at much less current.

As Russell mentioned, you can put a resistor accross the LED to split the SSR drive current between the resistor and the LED.  With a total of two resistors, you can set the SSR drive current and the external LED current separately.  Russell made it sound a bit mysterious, but it is actually very easy.  Here is a circuit:



Let's say we're using a common green LED for the external indicator.  As I said above, figure about 2.1V drop for one of those.  The max drop of the SSR LED is 1.4V, so that leaves 5V - 2.1V - 1.4V = 1.5V accross R2.  This resistor sets the SSR current, which we want to make 20mA in this example.  1.5V / 20mA = 75&Omega;

D1 will have about 2.1V accross it roughly independent of the actual current we put thru it.  R1 will therefore pass 2.1V / 150&Omega; = 14mA.  That's how much of the total current will NOT run thru D1.  The D1 current will therefore be about 6mA, which is plenty to light up most normal LEDs for indicator use.

Anyway, the point is not these particular values, but how you can use two resistors to set the currents thru both LEDs independently but still predictably enough for this application.  

Note that you can use the same idea if you want the indicator LED to take the higher current. In that case you put the parallel resistor over the SSR's LED. This could be useful if your indicator LED has to be visible in daylight and the SSR's LED has limited current capability.