Electronic – How does the phone detect if 3.5 mm jack circuit is closed

Sounds like you should consider placing some good quality capacitors in series with the audio signal lines from the jack that you added and to where you connected the inputs to the guts of the old cassette player's amplifier input. I would start with a value of 10uF with a voltage rating of 16V or more. It would be best to utilize non-polarized capacitors but if you cannot find those then consider one of two options:

a) Use two 22uF capacitors in series wires + to + and then one - lead to the jack and the other - lead to the cassette input.

b) Try to see how a polarized cap would work. Put the + lead toward the cassette player amplifier and the - to the jack.

These capacitor ideas block the DC bias that is apparently present at the connection point where you attached the signal wires into the cassette amplifier board. Some of the earlier devices that you connected to the cassette jack may very well have had capacitor coupled outputs. On the other hand your newest device may not have an output configured like this.

Edit: Actually, I found out the answer I posted is already given in the question you rejected as "too complicated". I ask you to reconsider that judgement, as anything involving operational amplifiers (suggestions also found in that question) will be even more complicated.

While the commentors are right that your circuit is a really bad idea in practice, your thoughts that lead you to the that circuit are actually quite good. You notice correctly that the voltage you obtain at typical volumes is too low to turn on the transistor (you need around 0.6V to get a significant effect). Using a darlington transistor doesn't help, as you experienced, because you need even more voltage to turn on a darlington transistor, but you can get away with a lot lower current. You problem is that your voltage is too low, not that your supply can't deliver enough current, so forget about darlingtons.

Increasing the "baseline voltage" as you call it, is exactly the way to go. Electronics people often call this process "adjusting the DC bias" of a transistor amplifier. It can be done quite easily on your circuit without adding a dangerous low-impedance power source like a battery. Let's work on your circuit. First, add a current limiting resistor into the LED circuit and put the LED on the other side of the transistor. It makes no difference in your circuit as it is now, but it will make soon:

^{simulate this circuit – Schematic created using CircuitLab}

You can fix your problem, as you correctly identified, by providing DC bias (a baseline voltage) to the transistor base. But you should not force a DC voltage into the headphone jack, so you need a capacitor for DC decoupling. This capacitor should pass the AC music signal, but block the DC required to bias the transistor. With the circuit as drawn, you can get the required DC bias (around 0.45V) from the battery using a voltage divider like this:

^{simulate this circuit}