Physical electron mass: renormalized VS running VS pole of propagator

Answers in some of the linked questions do contain the essential information that you need, but I think some conceptual clarifications may be helpful to you. I hope I've understood your question properly, and apologies if I say a lot of things you already know.

First, to set expectations properly, I'll only discuss on-shell renormalization. This is, as the name implies, a nice prescription when you can measure the on-shell particles directly. By the same token, it's not a good prescription for the case of light quarks, for example, which we never detect directly as free particles. Light quarks do of course have mass parameters which are physically important, but one needs to use some more abstract prescription such as $overline{MS}$ to deal with them, and the corresponding renormalized mass parameters don't have a clean physical interpretation in other more "familiar" terms.

But since we are considering on-shell renormalization, note that the physical mass that you measure with your detector is always at the scale $p^2=m_mathrm{ph}^2$ by Lorentz invariance. It doesn't matter what the energy scale of the process you're observing is. When that final state particle emerges from the process, it's on-shell. It always satisfies $p^2=m_mathrm{ph}^2$. You could boost to a frame where that particle is at rest if you want, whatever, it doesn't matter. In the context of an on-shell scheme, it simply makes no sense to ask if the "physical mass would run", or any similar such phrases.

So in on-shell renormalization the procedure goes like this. I compute the 1PI self-energy corrections to the propagator, sum the geometric series, and end up with a propagator like $$frac{-i}{p^2-m_0^2-M^2(p^2)}$$ where $m_0$ is the bare mass parameter in the Lagrangian, and $M^2(p^2)$ is what I got from the 1PI diagrams. Then I measure the physical mass with my detector and find it to be $m_mathrm{ph}$. Finally, I demand that $m_0^2+M^2(m_mathrm{ph}^2)=m_mathrm{ph}^2$, which will involve absorbing some infinity into $m_0$ plus an appropriate finite part. That will ensure that the propagator has a pole at the physical mass.

But the propagator is now so much more than just a pole at the physical mass. It's an interesting function of $p^2$, and you know that when used as an internal line in a diagram, it will have God knows what value of $p^2$ flowing through it. $p^2$ might even be negative. At these unholy values of $p^2$, $M^2(p^2)$ is producing all kinds of interesting effects in your propagator. This, in a sense, is the running mass, if you want to call it that. But again, it doesn't affect what you measure as the physical mass.

In fact, what I said above is not exactly correct, because $M^2(p^2)$ is in general complex. What I really meant is that $m_0^2+mathrm{Re}~M^2(m_mathrm{ph}^2)=m_mathrm{ph}^2$ should be true. Via the optical theorem, one can show that $M^2(p^2)$ will develop an imaginary part for values of $p^2$ for which some decay of the particle is possible (which depends on the details of your theory). We see that this imaginary part also affects the structure of the propagator, and in fact, is the reason why particles have a width in $p^2$ that is inversely proportional to their lifetime.