you've got 2 nanosecond edges inside the Remote.
2nanosecond edges are so fast, they serve as FINE IMPULSES to most circuits.
Thus the AM radio circuits are getting hit with tiny lightning bolts, and ringing, and you hear that.
"its safe to say they do not contribute with any EMI" though clearly the impulses do contribute, because the activity can be heard. An AM radio with 10KHz bandwidth (double sideband) has noise floor of -174dBm/rootHz + 10dB Noise Figure in frontend transistors + 40dB boost in noise floor from noise power being proportional to bandwidth, = -174 + 50 == 124 dBm. With 0dBm across 50 ohms being 0.632 volts PP, and -120dBm being 1 million times lower in voltage, the floor of detectability is about 0.6microVolts. Or 0.0000006 volts; now you want to bet on 5 volt MCU logic transitions NOT being detected by an AM radio, these receivers being notorious for static susceptibility.
So now we have some science, some actual math and physics, behind why the IR REMOTE can be detected by an AM RADIO. Neat, eh?
Now for some details on coupling between the IR Remote and the AM radio:
The remote will have several centimeters of PCB trace from the MCU to the LED driver transistor, which spits out 0.1amp or 0.2 amp currents for the LED, limited by 5 ohm or 10 ohm resistor. Into the transistor base will be 10mA with 2nanoSecond edges. From the collector will be 100mA (SWAG) with fast fall and slow rise (as the transistor exits saturation slowly). These currents may magnetically couple into ANY circuit loop inside the AM radio.
However, lets just think about capacitive coupling.
The AM radio is of non-zero size and we will assume several centimeters of PCB trace that are capacitively coupled to the IR remote.
So lets model these PCB traces: 2cm long 1mm wide, 2cm apart.
C = Eo * Er * Area/Distance = 9e-12 Farad/meter * 1(air) * (2cm * 1mm)/2cm
C = 9e-12 * 1mm = 9e-15 ~~ 1e-14 farads. [this ignores fringing & alignment]
Now lets compute a displacement current (the current generated by charging and discharging, by changing the electric field flux), between IR remote and
the AM radio.
Q = C * V; and we differentiate to get dQ/dT = dC/dT * V + C * dV/dT
now assume constant C (thru the air) and we have dQ/dT = C * dV/dT = Icurrent
Our injected (by changing electric field) current is
I == 1e-14 Farad * 3 volts / 2 nanoseconds
I ~~ 1e-14 * 1/nano == 1e-5 amp = 10 microAmps injected into the AM radio
Assume the impedance of the node is 1,000 ohms. Use Ohms Law, and you get
10uA * 1Kohm = 10 milliVolts.
And either the AM tuned circuits can ring, with this 2 nanosecond impulse,
or a higher harmonic (per Bimpelrekkie) may enter thru the antenna.
================== Now for magnetic coupling ===========
2 nanosecond edges are plenty fast for skin-effect in copper planes to cause some magnetic shielding and thus attenuation of the induced voltage.
We'll assume there is NO attenuation by planes, and just compute the worst-case induced voltage in the AM radio circuits.
As with the Efield coupling, assume 2 centimeter spacing between aggressor and victim. And assume the victim (the AM radio) has 2cm by 2mm loop. And assume worst-case alignment.
The relevant equation (ignoring some natural-log terms for easy math) is
Vinduce = [ MUo * MUr * Area/(2*pi*Distance) ] * dI/dT
where we'll assume dI/dT = 10 milliAmps / 2 nanoSeconds
Using MUo = 4*pi*1e-7 Henry/meter and MUr = 1(air, copper,FR-4, etc)
Vinduce = 2e-7 * Area/Distance * dI/dT
Vinduce = 2e-7 * (2cm * 2mm)/2cm * 0.01amp/2nanoSecond
Vinduce = 2e-7 * 0.002 * 0.01/2nano
Vinduce = 2e-7 * 2e-3 * 1e-2 * 0.5 * 1e+9
Vinduce (I don't have a clue how big/small this will be, until math is done)
= 4 * 0.5 * 1e(-7-3-2+9) = 2e(-12+9) = 2e-3 = 2 milliVolts magnetic coupling