Thanks for your reply. I guess I may not fully understand your writing and you didn't fully understand mine. So I may as well write up an answer to see if that helps improve communication and make it more concrete.
You said this is an Arduino, that it is driving other things, and that you figure you may need \$1.0-1.5\:\textrm{A}\$ from the power supply. Also, you obviously have considered a relay, mentioning specifically the Songle SRD-05VDC. But by the time you add the external parts to operate the relay, you may be better off just using a PMOS:

simulate this circuit – Schematic created using CircuitLab
If \$SW_1\$ bounces, your Arduino may wind up getting locked up in the process of starting. But adding debouncing externally complicates things, as \$SW_1\$ and \$D_1\$ are actually supplying power and the current may be over an amp, you said.
I'm ignoring the minimum \$V_{CC}\$ of your Arduino. \$D_1\$ drops a voltage and \$V_{CC}\$ may not meet specs after that drop. You could replace it with an appropriately sized Schottky and try that.
Regardless, your MCU is responsible for getting \$P_1\$ activated quickly so that it can pull down on the PMOS gate and keep the power applied. Your MCU would also monitor \$SW_2\$, of course, and then wait out the necessary delay before releasing it's hold-down of the PMOS via \$P_1\$.
The following circuit provides the needed timing, externally:

simulate this circuit
Again, you use a momentary pushbutton, \$SW_1\$, to turn on the power. The circuit starts counting down time, then. Once your MCU is running, it activates \$P_1=1\$ and this holds the circuit active. You monitor \$SW_2\$ via \$P_2\$ (which you need to configure with an internal pull-up, if possible, and an external pull-up resistor, if not possible) for the shut-down command using software you write. When your MCU detects that there is a shutdown in progress, the MCU causes \$P_1=0\$ (or high impedance) and this then will allow the timer circuit to start counting back down towards a shutdown of the MCU. Then the power is completely removed and everything is back to the way it was, once the time-out takes place.
The MJE170 should be large enough to handle your load and I've set \$R_1\$ and \$R_2\$ to be small values based on the need for a lot of base drive current for \$Q_1\$. Probably could get rid of \$D_2\$. \$D_1\$ provides a path for current from \$C_1\$ when \$Q_1\$ turns off. But you might be able to get rid of that for your needs, too (forcing the charge to decay out via \$R_3\$ instead.)
You might say, "Well, I don't want all that external timing stuff." Okay:

simulate this circuit
You have a debounced switch and your micro controls the timing.