A 7805 voltage regulator needs at least 7V into the input in order to regulate correctly. If you are feeding the bridge rectifier with 7.5 volts then, due to diode volt drops (about 1 volt on light load and more on heavier loads) the input voltage to the 7805 will be below 7 volts. That's your main problem as I see it.
Here's what the diode volt drop looks like for a 1N4148 and please note that this will be similar on any silicon diode: -
A 7805 takes about 4 mA unloaded so, as you can see, the diode forward volt drop will be about 0.7 volts. Because there are two diodes conducting in series in a bridge, the total volt drop will be at least 1.4 volts and suddenly your 7.5 volt battery is 6.1 volts when connected to the 7805. If the loading on the 7805 causes a current of 100 mA to be taken, the diode volt drop will be 2x ~1V = 2V and your supply voltage to the 7805 is only 5.5 volts.
Maybe try searching for a low drop-out voltage regulator - you should be able to find one that works down to an input voltage of about 5.5 volts or maybe a shade lower. If your load current is in the order of 1A then you will likely have to use schottky diodes and a low drop-out regulator.
You should also use decoupling capacitors on both the input and output lines of the regulator - read the data sheet.