Is there a physical mechanism explaining why a gauge boson "generates" this property of local gauge invariance?
I think you have it backwards. Starting with the Lagrangian for fermions, if we require local transformation ($\psi \rightarrow \psi \prime = \psi e^{iq\phi(x^\mu)}$) invariance, then we have to change the derivative in the Lagrangian into a "covariant" derivative, which includes a field term $A_\mu$ with its own transformation property that's related to the original local transformation ($A_\mu \rightarrow A_\mu ' = A_\mu - \partial_\mu \phi(x^\mu))$. This field is the photon, which is generated in response to the local transformation invariance requirement. Notice I haven't said anything about "gauge" yet.
The photon field happens to have a gauge invariance, because we can add the 4-gradient of any scalar function to $A_\mu$ and there will be no change to the measurable E and M fields.
The reason I think the local transformations are usually called "local gauge invariance" is that if a physical system is invariant with respect to a global transformation, and it remains invariant when that transformation is considered locally, then it is said to have a gauge invariance.
I kind of hoped that there might be some understandable connection between the local gauge invariance (which is kind of abstract) and a gauge Boson like the photon which is a more accessible thing (although very abstract too), in more physical terms.
Unfortunately, AFAIK the photon field appears because the math says it must, there is no "physical" reason.
(Those of you critiquing this response, please be gentle, it's my first attempt at answering a question on physics.stackexchange, I'm hoping it's acceptable!)
