Solid state relay is not completely open when control inputs are not connected
Figure 1. From the datasheet we can see that the leakage current is 7 mA.

A 12 W lamp will draw \$ I = \frac P V = \frac {12}{240} = 50 \ \text {mA} \$ (using 240 V to make the maths easy). 7 mA is a significant current for this lamp.


  When control inputs are not connected there is a drop of around 180 V across the relay. Why is this?


That's the voltage drop across the internal snubber capacitor. You have 180 V across it and 50 V across the lamp.


  When I provide 5V to the control inputs, I see a resistance to 10K ohm at the outputs of the relay.


Be very careful with this. You are working with live circuits. You can't measure resistance on a live circuit and, even if the mains side is dead, you can't measure the resistance of a triac in this way.


  What should I do to control the light using this relay?


All you can normally do is add resistance or another lamp in parallel with the lamp. If you are feeling adventurous you can isolate the power, open up the SSR and snip one of the leads on the snubber. Most will be partially filled with potting compound making this impossible.



Figure 2. See BigClive's Electronically controlled LED lamps glowing when off which explains this well.