This type of opto-triac is mostly used in mains voltage applications. Due to the limited current capabilities it's often used as a driver for a triac which is the actual switching device. Your requirements are modest, so you won't need that, and you can use the opto-triac to switch your load directly. The opto-triac is a cheaper solution than an electromechanical relay then, so at first sight looks like a better choice.
An important difference between electronic and electromechanical switches, however, is that the latter have a very low on-resistance, while the former always will have a voltage drop when switched on. That's the on-state voltage mentioned in the datasheet. This can be up to 3V, which in a 230V application won't matter much, but if your supply voltage is only 24V AC that's more than 10%. Your load will probably work at 21V, but you'll have to check it.
Repetitive peak off-state current is the leakage current when the triac is switched off. 2\$\mu\$A is a safe value.
Holding current is the minimum load current the triac needs to remain on when the gate is no longer driven. For an average triac your 20mA may be a bit low, but again the opto-triac's 3.5mA is a safe value. (Besides, the gate will be continuously driven, so it's a moot point. It is important in four-component dimmers, where the diac gives a pulse to switch on the triac, after which the triac is on its own.)
Then there is the minimum trigger current. That's the minimum current you have to supply to the LED to switch the triac on, and we'll have to calculate the series resistor accordingly.
Where did you get that 38\$\Omega\$ resistor value? You need figures 3 and 4 to calculate the value for the LED resistor. Figure 4 shows that 10mA is a safe value, and figure 3 shows that at 10mA the LED voltage will be maximum 1.3V. So \$R=\frac{3.3V - 1.3V}{10mA}=200\Omega\$ maximum. Your 38\$\Omega\$ would result in more than 50mA, which is not only more than Absolute Maximum Ratings (page 4), but also more than your microcontroller will be able to supply. So don't exaggerate, and pick a 180 \$\Omega\$ resistor. At lower resistances the current may become too much for your microcontroller's output. If you want more current through the LED (no more than 20mA, never use the Absolute Maximum Ratings!) you may want to use a transistor. Since you'd need a lot of them, consider a driver IC like an ULN2803.
In conclusion I think this opto-triac is a good choice. Alternatively, you may have a look at the MOCxxx series, for instance the MOC3012 needs only half of the LED current, which your microcontroller would appreciate. It doesn't give a nominal value for triac current directly, but from maximum power dissipation (300mW) we can derive that this should be 100mA. (It says peak repetitive surge current is 1A, 120pps, 1ms pulse width.)
SSRs use triacs as switching elements, and those are for AC, not DC. A triac is like two transistors so arranged that, once ignited, they keep each other switched on. The only way to switch it off is to reduce the current to below the hold current. AC devices do this 100 or 120 times per second, at zero crossings of the mains, but at DC it will not switch off.
Instead of interrupting the circuit you can also briefly short-circuit the triac's terminals to switch it off.
Best Answer
The specification is saying the LED will drop a voltage of 1.5V at a current of 50mA. However, you do not want to drive it that hard as any variance will kill the LED.
You need to deliver the rated current in order to ensure the device operates. The numbers in the spec sheet are a little weird here, but recommended led current is 5-10mA so lets pick 7.5mA.
simulate this circuit – Schematic created using CircuitLab
So you need to drop the micros output voltage down to 1.5V at 7.5mA, that means you need a series resistor of \$(3.3-1.5)/0.0075 = 240\Omega\$
Lets double check that to be sure. If the LEDs actual voltage is 1.16V, the current through the \$240\Omega\$ would then be \$(3.3-1.16)/360 \approx 9mA\$ which is still within the recommended values.