is it possible to use Laplace transforms on BJTs?
Yes, if you linearize the BJT about an operating point, i.e., do the s-domain analysis of the circuit with the BJT replaced by its small-signal model.
However, I don't think that's what you want in this case. This is non-linear, large-signal oscillator and so, you must do a large-signal analysis in the time-domain.
What you essentially have with this circuit is a relaxation oscillator.
Consider the initial condition that C1 is uncharged and Q1 / Q2 are off.
The voltage at the Q2's collector is zero volts and, since C1 is uncharged, the voltage at Q1's base is zero volts which is consistent with Q1 / Q2 off.
The capacitor will charge via current through R1 and, after some time, the voltage on the base of Q1 will increase to the point that Q1 begins to conduct which will, in turn, cause Q2 to conduct.
Now, we have positive feedback since, as Q2 starts to conduct, the voltage at Q2's collector will begin to rise which will add to the voltage at the base of Q1. This will act to further turn on Q1 and Q2 and, in a very short time, Q2 will saturate.
At this point, the Q2's collector voltage is fixed at about 2.8V and the capacitor current reverses; the path being "down" through Q2, through the capacitor, into the base of Q1. This acts to build a voltage across the capacitor that subtracts from the 2.8V at Q2's collector.
After some time, the capacitor voltage increases to the point that Q1 begins to turn off which acts to begin turning off Q2 and thus decreasing the voltage on Q2's collector.
Once again, we have positive feedback since, as voltage on Q2's collector begins to decrease, the voltage on Q1's base is decreased further which accelerates the turning off of Q2 and, in short order, both transistors are off.
Now, C1 can once again begin charging through R1 and the cycle repeats.
Below are plots of the output voltage and the voltage across C1 (positive referenced to Q1's base). Note that when the slope of the voltage is positive, the current is "left to right" through C1 and, when the slope of the voltage is negative, the current is "right to left" through C1.
Do you see a way to show this computationally? Hint: There are essentially two different regions of operation: (1) both transistors off and (2) both transistors on (Q2 saturated).