Why don't I need a resistance when testing a light bulb circuit in a breadboard?

With LED's, a small increase in voltage will result in a large increase in current. So it is really hard to get just the right voltage to keep an LED at the right brightness. If you let the voltage just get a tiny bit too high it may destroy the LED.

What makes it even harder is that as the LED gets hot, the current will also increase. Naturally when you power it up it will tend to get hot. As a result it is just too much trouble to drive an LED with a voltage. Some form of current limiting usually has to be put in place. It doesn't have to be a resistor, but that is probably the most simple way to do it.

LED light bulbs have circuitry integrated into them that overcomes all these problems.

Old-fashioned incandescent light bulbs (including halogen bulbs) are different. The part that lights up is made from a thin tungsten wire that glows when it gets hot. The wire has resistance which limits the current automatically. This resistance is also what causes it to heat up. And, icing on the cake, the resistance goes up with temperature, so incandescent light bulbs are really stable when powered from a voltage source.

The "problem" of an LED (and each diode) is that it has not constant (static, ohmic) but "dynamic" resistance that decreases when the voltage applied to the diode increases. Thus, in Ohm's law, both voltage and resistance change but in opposite directions - I = Vinc/Rdec. As a result, the current through the diode vigorously increases... and if the applied voltage is quite higher (9 V here) than the diode threshold voltage (typically 2 V for an LED), the current and accordingly the power, will become unacceptably high. To solve the problem, we connect a resistor in series. The operation of this network is visualized in Fig. 1.

Fig. 1. LED presented as a voltage-stabilizing dynamic resistor

When we turn on the power supply, the (input) voltage VIN increases from zero to maximum. In the graphical representation, its IV curve (including the resistance R) moves to the right (translates). At the same time, the diode begins decreasing its static resistane RST so its IV curve rotates counterclockwise. As a result, the operating point A slides up along the vertical part of the diode IV curve. The current variations are significant while the voltage drop VD (VF) across diode does not change - the diode differential resistance is zero.

The incandescent lamp also does not possess constant resistance. However, in contrast to an LED, its dynamic resistance increases when the voltage applied across the lamp increases. Now, in Ohm's law, both voltage and resistance change but in the same direction - I = Vinc/Rinc. As a result, the current and accordingly the power, more slowly increase... and will not become unacceptably high.

When you say lightbulb I take it you mean an incandescent light source? If so, the 9V battery has nowhere near enough juice to do any damage to its filament, which is usually made out of carbon, tungsten, or titanium. Two of the filament's requirements are high resistivity and high melting point, which is one of the needed features to make it glow and produce light. You might even say, it is its own "resistor"...