Why is inductive kickback not V=IR?
To me, keeping the current flowing exactly as it was means the current is the same.

The inductor &quot;tries&quot; to keep the current constant, or &quot;does everything in its power&quot; to keep the current constant. That doesn't mean that it actually keeps the current constant.
Similarly, a resistor &quot;resists&quot; the flow of current. That doesn't mean the current through a resistor is always 0.

The inductor therefore acts as an instantaneous current source.

Correct. If you implement what SPICE calls a transient analysis, you will find that at each time step, the inductor is treated as a current source (representing this tendency to keep the current constant) in parallel with a resistor (representing the possibility, if the inductor is not shorted, of the current changing infinitesimally before the next time step).

Since everyone says that \$V=L\frac{dI}{dt}\$, I'm inclined to think that it is correct, but I'm struggling to understand why \$V=IR\$ is wrong on this occasion.

An ideal inductor doesn't have any parameter &quot;\$R\$&quot;, so there's no way you could write the equation \$V=IR\$ to describe the ideal inductor.
\$V=IR\$ is an equation that describes the behavior of an ideal resistor.
\$V=L\frac{dI}{dt}\$ is an equation that describes the behavior of an ideal inductor.
In some particular circuit, \$V=IR\$ might well describe what is happening to a resistor that is connected in parallel with an inductor. But it can't describe what's happening to the inductor itself, because an inductor is not a resistor, and doesn't even have a parameter \$R\$ to determine its voltage or current.

I've drawn a circuit below. In this case, I would expect the voltage over the inductor/resistor after switching the current source off to be V=IR=100 V.

This is not correct. If \$V\$ is the voltage at the terminal where the resistor and inductor are connected, but not the one that is designated as ground, then for current to continue flowing top-to-bottom through the inductor, the resistor current must be flowing bottom to top. And thus the voltage must be -100 V, not +100 V.
Which is a good thing, because if it were +100 V, then the inductor current would be increasing rather than decreasing, and the whole circuit would be violating conservation of energy (since the resistor and inductor would both be absorbing power but nothing would be providing power).