Violation of Stefan's law when shining a light on a black body

Stefan-Boltzmann law

The Stefan-Boltzmann law says quite simply that the thermal power $P_text{thermal}$ emitted by a black body at temperature $T$ is $$ P_text{thermal}(T) = A sigma T^4$$ where $A$ is the object's surface area and $sigma$ is a constant $$ sigma equiv frac{2 pi^5 k_b^4}{15 c^2 h^3} , .$$ This law is always true as long as the black body can be said to be at a certain temperature.

Addressing the original question

Now a ray a light is shone upon it. Since a black body absorbs all radiation that falls upon it, it must absorb more radiation than it did when no light fell on it...

Correct so far.

but according to Stefan's law the radiation absorbed depends only on the ambient temperature (which is constant).

That's not correct. The Stefan-Boltzmann law says that the power emitted by a black body depends only on its temperature. The power absorbed depends on how much radiation is coming onto the black body, and that radiation does not necessarily have to be coming from an environment in thermal equilibrium. The incoming radiation can be of essentially two different types:

1. Thermal radiation, i.e. radiation emitted from objects that various other temperatures ("objects" here include space, actually).

2. Coherent radiation, i.e. light from a laser.

To some rough approximation, most natural electromagnetic radiation has properties that put it somewhere between fully thermal and fully coherent light $^{[a]}$.

Suppose the black body is initially at temperature $T_i$ and is sitting in a room, also at temperature $T_i$ with no light sources. In this case, the object is at thermal equilibrium so the power absorbed and emitted must be the same. We know from the Stefan-Bolzmann law that this power is $$ P_text{thermal (absorbed and emitted)}(T) = A sigma T^4 , . $$ If we now turn on a light source (e.g. a laser) with power $P_text{source}$ and point it at the body, then the total power flow into the black body is $$ P_text{in} = underbrace{P_text{source} + P_text{thermal}(T_i)}_text{absorbed} , , underbrace{- P_text{thermal}(T_i)}_text{emitted} , . $$ At first, the emitted and absorbed thermal powers are equal, so they cancel and the total power into the black body is $P_text{source}$. Since the black body is absorbing more than it's emitting, it's temperature starts to go up. Therefore, by definition, the system is no longer said to be in thermal equilibrium.

Suppose the black body absorbs enough power that its temperature rises to $T$, while the surrounding environment is so big that no matter how much energy it absorbs, its temperature is still approximately $T_i$. Then the power going into the black body is $$ P_text{in} = P_text{source} + P_text{thermal}(T_i) - P_text{thermal}(T) , . $$ This system is never truly in equilibrium because we're constantly adding energy through $P_text{source}$, but there is a "steady-state" where the temperatures of the objects stop changing. That happens when $P_text{in} = 0$, i.e. when the total incoming and outgoing power from the black body are equal, i.e. when begin{align} P_text{in} &= 0 underbrace{P_text{thermal}(T_f)}_text{emitted} &= underbrace{P_text{source} + P_text{thermal}(T_i)}_text{absorbed} end{align} where $T_f$ is the final steady state temperature of the black body.

Homework: Solve for $T_f$ in terms of $P_text{source}$ and $T_i$.

Side note

By the way, pretty much every single idea about "breaking the second law of thermodynamics" or making a "perpetual motion machine" comes down to someone misunderstanding that the theoretical results of equilibrium thermodynamics only apply in equilibrium.

[a]: I'm intentionally not explaining that in full detail because it would take us too far away from the main point.