Why does this circuit cut out when an extra load is added?
Your motor is a lot less sensitive to voltage than your circuit is. You are probably looking at this:


But front-page "advertising" numbers can't be trusted, especially for numbers with more complicated behaviour. You should be looking deeper in the datasheet at this:


Your battery's internal resistance is 0.12Ohms, at best. At 1.5A you lose 0.18V per cell. If your nominal voltage is 1.5V per cell, then you are 5.28V.

If your battery is brand new, the internal resistance is 0.2Ohms and you have 4.8V when drawing 1.5A if using the nominal voltage. But your brand new battery probably starts at a high enough voltage to counteract this and more so your initial voltage is higher than nominal rather than lower.

Your LM7805 has needs a headroom (has a dropout voltage) of 2V on top of its output to operate when supplying 1A. You are drawing less current than that so the dropout voltage is less but you are still probably operating the regulator in a region where it's just struggling to pass all the input voltage straight to the output but it's a regulator, not a switch so even more voltage is dropped in that process as well.

According to the datasheet of the LM7805CV manufactured by ST at 220mA, the dropout is ~1.6V.