Might it work?
Yes, if you use components with appropriate values.
Could it put some impedance load on the AC line or disturb it somehow?
Yes, but unless you use an unnecessarily large transformer core, the load will be very small.
In your diagram, you show more turns in your primary winding than in your secondary. Almost certainly, you would want this the other way around. That is, you want the transformer to step-up the voltage. Also, I would put two LEDs (or one LED and one vanilla diode) in the circuit in anti-parallel configuration. In parallel, but with opposite polarities. That will help to keep the secondary conducting when the power is on. With only one LED, when the secondary stops conducting, there is typically a voltage spike. If your transformer has a sufficiently large core, that spike could damage the LED.
The turns ratio, and the core size will depend upon the current flowing through your power line, the current you want to drive your LED, and whether you want to run your transformer in linear mode (avoiding core saturation) or saturated mode. For most applications, one chooses a transformer core that is big enough to avoid saturation. Saturation introduces non-linearity, and in the case of a transformer that is driven by a sinusoidal voltage source, it can lead to excessive currents.
However, in your circuit, core saturation may be your friend. In a transformer driven by a sinusoidal current source, saturation can serve to limit secondary voltage (and consequently current). Further, if your concern is simply detecting current in a power line, rather than measuring it, non-linearity can allow a secondary output that is relatively constant over a wide range of primary current levels.
First, how to determine whether a core will go into saturation. If you have a current transformer with a single turn primary, the magneto-motive force \$F_m\$ is just the current.
\$F_m=NI=I\$
H is given by
\$H = \frac{\displaystyle F_m}{L_m}\$
where \$L_m\$ is the effective length of the magnetic circuit.
Then we will deny reality, and assume that there is a constant \$\mu\$ that relates \$B\$ and \$H\$ so:
\$B=\mu H\$
Putting it all together, the core will saturate if
\$\frac{\displaystyle \mu I_{max}}{\displaystyle L_m} \gt B_{sat}\$
Now, if you are operating in linear mode, (no core saturation), then the ratio between the current through your power line and the current on the secondary side is just the turns ratio.
If, for example, the current through your switch is 10A, and you want your LED to run on 20mA, then your turns ratio needs to around 1:500. (Of course unless you use a bridge rectifier, the LED will conduct only half the cycle so your average LED current will be 10mA). If the current through your switch is only 50mA, because, say it is only driving a relay, and you wanted to run your LED with 20mA, you would use a 2:5 ratio.
No suppose you want to saturate your transformer core. In that case, at each half cycle, \$B\$ swings between \$B_{sat}\$ and \$-B_{sat}\$, and the rest of the time, \$B\$ stays constant. When \$B\$ swings between its extreme values, there is an EMF (i.e. voltage) pulse generated in the windings which has a Volt-Second value of
\$ET = 2NB_{sat}A_{cross}\$
where \$A_{cross}\$ is the effective cross section of the core. There will be one positive going pulse and one negative going pulse per cycle. If the mains frequency is f (probably 50Hz or 60Hz) then the average voltage of, say the positive pulses only would be
\$V_{avg} = 2NfB_{sat}A_{cross}\$
Notice that the equation does not depend upon the current through the power line. Once core saturation is reached, the \$ET\$ (volt-seconds) of the pulse is independent of any further increase in current. This, may be an advantage in your application. However, the width of the pulse, will narrow as the power line current increases. That is, higher currents will generate shorter pulses of higher voltage. This may necessitate the use of an RC or LC filter to smooth the pulse.
So, with a given core effective cross section, core effective length, \$\mu\$, \$B_{sat}\$, and knowing that the core is saturated, we can pick a resistor, (possibly with smoothing capacitor and/or inductor) that will drive the LED with a smaller sized core, and without precise knowledge of the current in the power line.
