Let's take this in two parts: first an exhibition of how this works for a trained physicist who has access to the tools of multivariate calculus, and second an examination of how you might explain this to students in an introductory class based on algebra and trigonometry (no calculus).

Sophisticated view

Just as the proper formulation of the Newtonian dynamical rule is $\vec{F}_\text{net} = \frac{\mathrm{d}\vec{p}}{\mathrm{d}t}$ rather than $\vec{F}_\text{net} = m \vec{a}$, the proper formulation of the dynamical rule for rotations is (treating the fized axis case so we can dispense with vector notation): \begin{align} \tau_\text{net} &= \frac{\mathrm{d}L}{\mathrm{d}t} \\ &= \frac{\partial L}{\partial \omega} \frac{\mathrm{d}\omega}{\mathrm{d}t} + \frac{\partial L}{\partial I} \frac{\mathrm{d}I}{\mathrm{d}t} \\ &= I \alpha + \omega \frac{\mathrm{d}I}{\mathrm{d}t}\;. \end{align} Of course, in the case of rigid objects in free rotation we have $\frac{\mathrm{d}I}{\mathrm{d}t} = 0$ so that this becomes $$ \tau_\text{net} = I \alpha \;,$$ but for mutable objects or cases where the axis of rotation is moving we need both terms.

Further when the net external torque is zero we can write $$ I \alpha = -\omega \frac{\mathrm{d}I}{\mathrm{d}t} \;. $$

Classroom view

The students don't have the mathematical tooling to parse the argument above in the form written, so we need to provide a scaffold of some kind.

Work it as a conservation problem

(Following a suggestion from Acccumulation in the comments.)

If you have the rule of conservation of angular momentum, you just go with \begin{align} L_f &= L_i \\ I_f \omega_f &= I_i \omega_i \\ \end{align}

Introduce the idea that you need a term for changes in the inertial tendency

I took a crack at this in the comments, and as you say it is less than satisfactory because those "dumps a load straight down" problems really involve multiple parts of a system in a way that is not exactly analogous to the question at hand.

You're clever student is likely to land right on the difference if they are presented together.

Tackle it at the level of forces and torques to motivate the need to a second term

(This is what you asked for in your follow-up comment.)

The key observation here is to track a single mass element through a change in radius from $r_1$ to $r_2 \ne r_1$.¹ During the time that radial change occurs the objectmass continue to turnmove "around" the center of rotation, but the path of the mass element is not a circle centered on the axis. That means that the net forces acting on the mass element are not centripetal, and therefore $\vec{r} \times \vec{F}_\text{net} \ne 0$! Even though the forces are internal, they can exert a toquenon-zero work on the mass element.¹

now to result in an overall change in angular velocity we require that the 3rd law pair force acts on a mass element opposite² the one we considered initially so that even though $F_{a,b} = -F_{b,a}$ the toques do not cancel each other out: $\tau_{a,b} \ne \tau_{b,a}$$\vec{F}_\text{net} \cdot \vec{s} \ne 0$.

Have the student(s) check this themselves.

¹ This is also how internal forces can do work on the body allowing But that makes the rotationaltranslational kinetic energy of the mass element increase when coming closer to change duringthe center or decrease when moving outward. Either way there is no way for $\omega$ to remain constant.

None the less for central forces we still have $\tau = 0$. But that leads to a angular momentum-preserving deformationcontradiction if you insist that ($\vec{F}_\text{net} \cdot \vec{s} \ne 0$, as well)$\tau_\text{net} = 0$ is the complete rule for these system. As a result we must introduce a part that depends on varying $I$.

²¹ Strictly it doesn't have to be exactly oppositeThis is completely natural in the orbital problem that you offer, but use this case to make the checking easyis worth saying explicitly so we'll remember when we work on mutable solid objects in rotation.