Voltages don't "come from" someplace, they exist between two nodes. The notion of a "foreign" voltage is non-sensical. In the context of electronics, you can't talk about a voltage in absolute terms, unless it is implied what other node the voltage is relative to. If the circuit has a node labeled ground, then it is understood that voltages talked about as being absolute are really relative to ground. That's actually the main purpose of labeling a node as ground in a schematic.
Your circuit doesn't show any node as being the implied ground. However, from context and the way the schematic is drawn, we would usually assume that the implied 0 V reference is the negative lead of the power supply and the cathode of the LED. Forcing people to make assumptions like this is a bad engineering, so ground should really be clearly marked if you're going to talk about voltages on single nodes without explicitly mentioning the reference node.
The transistor only reacts to voltages between its pins. To turn on a transistor, you have to put enough voltage on B with respect to E so that sufficient base current flows. Since B-E looks like a diode to the outside circuit, it is usually better to specify the B-E current instead of the B-E voltage. The voltage will usually be 600-750 mV over the useful operating range, but the current will vary much more widely over that range.
The circuit as you show does nothing because the negative end of V1 is not connected to anything. You can't say that it's putting 5 V on the base of the transistor, since that voltage isn't referenced to anything in the transistor's circuit. With the negative end of V1 open, current can't flow thru V1, so it becomes irrelevant.
If you were to do this with a real power supply, it would have some capacitive coupling back to the power line and the transistor circuit. Connecting the positive end of V1 to the transistor base would couple some capacitively picked up noise onto the base that is ultimately referenced back to the transistor circuit. If the transistor were driving a speaker, you'd probably hear some hum. Unless you have a very unusual situation, the part of the capacitively coupled signal referenced back to the transistor circuit won't be strong enough to light the LED.