Firstly, the feed to the cameras:
I think, to be safe you should use a low drop-out regulator to feed your cameras - this takes care of slight overvoltages. The KA278R12C is a linear voltage regulator with very low drop out: -
Note that even when the input voltage is at 10V, the device is still able to ostensibly produce 10V at its ouput when delivering over an amp (6 ohm load). I suspect this device will be good enough to feed your camera system but I can't absolutely say because you haven't specified current. There are other higher power devices that would fit the bill.
Can I wire a load to my battery if it is connected in parallel with the charger?
If the battery is lead/acid and the charging current is significantly more than what the camera load takes when attached to the above regulator then yes you can. If the battery isn't lead-acid then we need to know which technology it is.
How can I add a solar panel + controller to the previous circuit?
Playing safe, you can use a relay circuit that activates the relay when the AC power is applied to the charger - the relay contact can switch the battery from solar charger to AC charger in a few milli seconds. Playing a little bit unsafe, it's likely that your solar charger will have a diode in its output that protects the battery from discharge when the sun doesn't shine.
This very same component probably can mean that you can connect the AC charger permanently to the battery (and solar charger) BUT, you may need to add a series diode\$^1\$ in the AC charger's output when AC is off and the solar charger is feeding juice to the battery; the AC charger's output circuits may be activated by the solar charger and it's difficult to say what will happen - worst case it might pop the output transistor in the AC charger - best case no problem.
However, the chances are likely that your AC charger (just like your solar charger) will be protected from reverse voltages when power is down (or sun is not shining). You need to check this.
\$^1\$The diode needs to be a low volt drop schottky type capable of taking the charge current (again, you haven't specified max charging current so it's impossible to say but there are plenty rated for 10A and 20A continuous usage).
If you want NEVER run out of battery, you should come from ALWAYS extreme cloudy condition at shortest day, because weather statistics cannot guarantee cloudy time duration limits.
Note, solar battery manufacturer gives power output for very optimistic conditions, say clear day and battery oriented directly to sun.
Daylight wikipedia article gives you range of conditions you'd expect. According to it, in extreme cloudy day there's less than 0.002 of max. light intensity. Add here 40% daylight time. So you will need rated power of ~2000 times consumed to work forever (1 KW solar battery for 0.5 W device).
So you have to define less strict requirement to get reasonable values, say 10% per year failure rate. Then I expect about 10-20× solar power, say 5-10 W solar panel and larger accumulator (week of autonomous power). But accurate value require much statistical calculations or experiments.
Best Answer
I'm going to assume that your router uses about 7W, as measured on a similar router.
If you don't consider inefficiencies then the math works like this:
Nominal battery power rating: \$12 V \times 30 Ah = 360 Wh\$
Dividing by your consumption: \$\frac{360 Wh}{7 W} = 51.4 hours = 2.1 days\$
But there are inefficiencies. I assume you are connecting the battery to an inverter, then the inverter to the router's power supply. Each of these conversions wastes energy. If each conversion is 85% efficient, then you get a combined inefficiency of \$0.85 \times 0.85 = 0.7225\$. This means that the entire system is 72.25% efficient.
So, multiplying it out: \$51.4 hours \times 0.7225 = 37.1 hours = 1.55 days.\$
This should be fairly realistic, although you may want to verify the values (router power, inverter inefficiency, etc)
Note that this figure is simply based on a full battery. It doesn't consider the amount of power provided by the panels. We would need a lot more information to try to predict the power generated by the solar panel.