3.7V 1500Ah battery charging issue
This circuit is quite wasteful of the energy in your four 3.7 V batteries.  A buck converter would be much more appropriate here than a linear regulator.

For 1 A out, your regulator would dissipate over 9 W.  That will require carefully designed cooling.  A buck converter would not only extend battery life significantly, but will be smaller and cheaper than this circuit after the heat dissipation issues are addressed.

A 7805 isn't stable without a cap directly on its output, not thru a diode.  See its datasheet for the details.  There should also be a cap immediately on its input.  10 &micro;F ceramic, not electrolytic, for both would be fine.  

I don't know what you think the diode is doing for you, but that's a bad place for one.  If you are afraid the battery might get below 5 V and you don't want power applied to the USB output (quite unlikely) to back drive the battery, put it directly in series with the battery.

Note that the integral diode of a buck converter would take care of blocking reverse current without additional parts.

Even if there was a reason for the diode, a Schottky would be more appropriate here.