Some help with 9V Longlife VARTA batteries

The transistor may act as a switch or a variable resistor. If no voltage is applied to the base (more precisely: no current flowing into the base) then the switch is open. As base current is applied it gets amplified by the transistor into an N times larger collector current. The "N" is an important transistor parameter, called \$H_{FE}\$, and it defines the current amplification factor. For general purpose transistors this is often around 100.

So if you apply a base voltage (you need a series resistor!) so that there will flow, say, 1 mA, then there will be 100 mA collector current if the circuit allows it. That means that other components may limit that current to a lower value. Let's assume your LED has a 2 V voltage drop, that will be rather constant for that type of LED. Then assuming the transistor is fully conducting (no voltage drop between collector and emitter) you'll have 9 V battery voltage - 2 V LED voltage = 7 V across the resistor. If we choose a resistor value of 350 Ω then, according to Ohm's Law we have a current of 7 V/ 350 Ω = 20 mA through that resistor and therefore also through the LED. (20 mA is a typical current for an indicator type of LED.)
So, while the transistor would like to draw 100 mA, the resistor will always limit that to the lower 20 mA.

You don't say what the signal from the amplifier is. Is that a line level (500 mV) or a speaker output level (3 V for 1 W)? In the first case the voltage will be too low; a transistor's base has to be at 0.7 V minimum before current starts to flow. If you use the speaker output you can use a 1 kΩ resistor in series with the output to limit the base current.

Also place a diode (1N4148) in anti-parallel with the base: cathode to the base, anode to ground. This prevents too large negative voltages across the base, which would destroy the transistor.

This sort of pop is a common problem with audio connections, particularly guitar leads as they can get swapped live. It's caused by capacitance in the systems being connected holding different dc levels, and results in a step pulse in the power output stage, which isn't good for your speakers and probably is what overloads your battery.

One means to neutralise is a reed switch in the jack, though I don't know of a mini stereo version.

The other means are to reduce capacitance in the cables and input stage of the amp. It may also be possible to use a transformer to isolate the input, or possibly a band pass filter to remove both the suddenness of the step change and the dc offset. But a filter may well have to be in the audio range to be effective, and that is the range you want to maintain.

If there was an automatic circuit which did this well then every electric guitarist would buy one, and there doesn't seem to be any out there (normally you have manual mute switches or the fore-mentioned silent jacks), so if you do get it to work there's quite a market.

An answer in another question pointed to this circuit as a possibility.