You are confusing first order approximations or rules of thumb with exact reality. Yes, the emitter voltage of a emitter follower with a fixed base voltage stays constant to first approximation, but when you're pushing the limits you have to look more carefully than that.
There are a number of problems with your circuit.
The zener voltage won't be exactly constant over its current range. Note that its current range will be considerable, since the base needs anywhere from 0 to about 20 mA over the 0-1 A output load range. You still need a little current thru the zener at full output. Let's say 1 mA. That means you need to feed it with 21 mA at no load. Can this zener even do that? 21 mA x 5.6 V = 120 mW, which is within the plausible range for some zeners.
The 2.2 kΩ resistor can't supply the necessary zener current. You have 20 V sine in, which is 28.3 V peak. Let's say 1.5 V lost in the full wave bridge, so you have 27 V max on the cap. (27 V - 5.6 V) / 2.2 kΩ = 9.6 mA, which is way too little to sustain 1 A of output current.
The transistor will get quite hot. If you were able to feed its base to keep the output at 5V, it would drop 22 V, which means it will dissipate 22 W. That's going to reqire a serious heat sink.
You could fix this circuit to provide more base drive and more active feedback to regulate the output voltage, but then you'd just be reproducing a 7805 regulator at best. Those have all that stuff built into a single 3-pin part, regulate the voltage well, can handle 1 A, and even have thermal shutdown to prevent damage in case of overheating.
However, what you really want is a switching power supply. That will be simpler and cheaper than having to deal with the heat from a linear regulator.