The output of voltage of the transformer will be proportional to the input voltage. If it is 15 VAC at 220 VAC in, then it will be 13.6 VAC at 200 VAC in, for example.
You seem to have a lot of headroom, so should be able to tolerate significant power voltage sag. In fact, you seem to have so much that your supply is quite inefficient. At 15 V out, the peaks of the waves will be 21.2 V. Assuming a full wave diode bridge, you loose about 1.4 V leaving 19.8 at the peaks. This is the level that the capacitor after the full wave bridge will be charged to twice per line cycle, which is every 10 ms for 50 Hz power.
The 7805 regulator needs about 7.5 V in to make reliable 5 V out. There is a lot of room between the 7.5 V minimum the regulator needs and the 19.8 V peaks the cap gets charged to. The cap voltage will drop between the peaks depending on the cap size and the current. Usually things are sized for a few volts drop. For example, let's say you have a 1 mF cap and the current draw is 100 mA. That would only sag 1 V over 10 mS, which brings the lowest voltage into the 7805 down to 18.8 V.
We can work this backwards and see what the minimum necessary line voltage is in this case. You need 7.5 V into the regulator minimum. The sag between peaks is still the same because it is only a function of the current draw and the capacitance. That means the cap needs to be charged to 8.5 V at the peaks, which requires 9.9 V AC peaks before the full wave bridge, which is 7 V RMS. That is 47% of the output at 220 V, so your minimum input limit is 103 V.
Of course you should always leave some margin because stuff happens, but any vaguely reasonable "220 V power" line isn't going to sag as low as 103 V unless something is very wrong.
As for another solution, get a "universal input" power supply. These are designed to work with any of the worldwide house power. They are usually specified for something like 90-260 V AC, 50 or 60 Hz. They are switchers, so are a lot more efficient than your linear regulator. This is really a better answer anyway than a big iron line power transformer type power supply.
To increase input voltage range of your supply with linear regulator you should use some step-down converter.
Thats the only reasonable method to "loose" more than 10-20V at 2A peak current.
Correctly designed simple step-down converter will decrease voltage with little power losses and heat generation.
MAX5024 has low current because you get 6W of heat at 100mA and 60V difference between in and out pin. Thats a lot. For 2A it would be 120W :) So forget about loosing 30-60V on linear regulator with 2A current requirements. Thats insane.
There are many integrated switching regulators (like easy to use LM2574 family), but I can't find any capable to deliver 2A output with 60V input. Maybe someone will suggest something.
Best Answer
Why bother with a zener....
The zener itself consumes current so you are in fact increasing the power usage not reducing it. Further, the ZENER would need to be a high current device so that the series resistor does not have to be so high that it affects the regulator. That's even MORE power to dissipate and lose.
If you MUST take some of the load off the regulator, and your load is 150mA, just use a suitable rated resistor to drop the 12V down to just above the minimum voltage required by the regulator at that current. Perhaps a 1W 30R. Or use a high current zener in place of the resistor.
It won't make it more efficient, and you will still have to get rid of as much heat, but it won't be worse and heat-sinking may not be required.