Assuming it's a "universal motor" which is the normal type of brushed motor running on AC, it will run equally well on AC or DC of either polarity, but always in the same direction. (As Olin says, the AC supply reverses polarity 100 or 120 times per second already. So, simply reversing live and neutral will not reverse the motor.
The trick to reversing these; and what sets them apart as more flexible than a permanent magnet motor, is that there are two windings : one on the rotor, and one fixed to the motor frame called the field winding. To reverse it, you need to reverse the connections to the field winding OR the rotor, BUT NOT BOTH.
If it is described as a reversible motor, these windings will be separately available as 4 external connections, possibly linked together, but in a way that you can disconnect one winding and connect it the other way round.
It's easy to identify the field winding : it has higher resistance and consumes much less power, maybe 10% of the total. The rotor winding goes to the brushes.
At which point a photo of the connection box is probably useful.
But assuming this is true so far, you can reverse the connections to one winding (usually the field winding which consumes less power) and test that the motor runs backwards - no circuitry needed so far.
Alternatively, the rotor winding may be accessible via screw terminals on the brushes.
Then to switch between forward and reverse you need something like a DPDT switch (hand operated!) or relay (remote operated) to reverse these connections on demand. It may be better to stop the motor, reverse it, then restart, to prevent extreme mechanical stresses and current surges.
As for your other question : the 1 amp current is generally the current at rated load :you can bank on the starting current being much higher (easily 5x, maybe 10x as much). To estimate it, measure the DC resistance of the motor - say 11 ohms - and divide 110 (or 230) by it (for 10A start current). This is also the "stall current" which the motor will draw if you stall it while powered. A "motor rated" breaker will allow stall current briefly, but trip if the current persists for long.
Your maths is wrong.
100 Watts generated by solar power at 12V.
Your 8.33 amps is the correct calculation.
The mistake you have made is assuming that this current stays the same value when added to the 110V source. You cannot just connect the systems. To feed the 12V (DC) into the 110V (AC) you need to convert them to the same thing - usually the 110V AC. This means using some form of converter (basically a transformer with the DC turned into AC by transistor switches). The POWER OUT will always be less than the POWER in so that at 110V the current will be less than 0.909 Amps NOT 8.33 Amps. (Vin * Iin >= Vout * Iout)
Best Answer
A few things come to mind:
You need to protect the bus from overcurrent. Even if your power supply has its own built-in protection, I would use an additional fuse (or circuit breaker) at the output of the power supply. This way you can be sure the maximum current won't get away from you. This leads to:
You'll need to use beefy wire. In the USA, 15A circuits are wired with #14AWG, minimum. If you want to use thinner wire, you'll have to fuse each leg appropriately.
If you do use #14AWG wire (or your local equivalent), don't use the typical cable used for household AC! Although it would work out technically, it would cause major confusion and ambiguity. You don't want anyone expecting 12VDC and getting Mains voltage (now or in the future).
At 12V, the current draw can quickly add up. Keep this in mind as you add devices. You may want to swap in a 24VDC power supply in the future. It is a common industrial standard, gives you twice the power over the same wires, and still falls into the "low-voltage" category.
Adding to Point #4: If you choose local 5V converters that accept a range of input voltages (including 12V and 24V, of course), then you won't have to change anything if you bump up the supply voltage.