wiring 555 to disconnect itself from electricity
Something like this will work. Timing and bypass components and load omitted for clarity. 







simulate this circuit &ndash; Schematic created using CircuitLab

When the output goes high, it turns on Q1 through R1, which turns on Q2 via R2. R2 should be low enough that Q2 stays reliably saturated with the highest load on the 555, plus the 555 supply current. R3 prevents leakage in Q1 from being amplified by Q2, so the total leakage in the passive state is merely the leakage in Q1 + Q2 (probably in the 10's of nA at room temperature), rather than tens or hundreds of times higher.  

Edit: Expanding on the leakage- suppose the leakage of Q1 is 20nA, and gain of Q2 at 20nA is 40. All the leakage current from Q1 flows into the base of Q2. Then the collector current of Q2 will be 40 * 20nA = 800nA plus another 20nA for its leakage, so a total of 840nA. That's without R3. 

Now add R3- the voltage across R3 will be 20nA * 20K = 400uV. That means virtually all the current will flow through the resistor and not the base (since the base-emitter junction doesn't really start to conduct until the voltage across it reaches hundreds of mV) so the leakage of Q2 is just the minimum 20nA, for a total of 40nA (21x better performance for a single resistor). 

Edit2: 
Further to discussing why use this circuit as opposed to lifting the GND pin, here's the schematic of the original 555. It should be reliable as it is from the designer's (Hans Camenzind, RIP) book Designing Analog Chips. 



Edit 2: 

In order for the trigger to work properly in conjunction with the above-discussed power switching scheme- consider the below schematic. R6 may not be necessary.







simulate this circuit