The anomalous dimension for the field strength is defined as (eqn 12.63 Peskin):
$\gamma = \frac{1}{2} \frac{M}{Z} \frac{\partial Z}{\partial M} = \frac{1}{2} \frac{\partial
\log Z}{\partial \log M} $.
This definition always holds. What you actually calculate for the right-hand side of the above equation once you have a Z within a particular scheme will be in general scheme-dependent.
Sorry, I can't help you with the $O(N)$ vector model...
There is a Ward identity that links the charge renormalization to the photon's wave function renormalization. Ward identities are relationships between correlation functions that follow from the quantum theory having a symmetry. In this case the gauge invariance of QED relates (among other things) the electron's two point function (propagator) to the $eA\bar{e}$ three point vertex.
If we write out the lagrangian including arbitrary scalings for $A$ and $\psi$ and I also put in a constant $Z_e$ to let the charge scale
\begin{equation}
\mathcal{L} =-\frac{1}{4} Z_1 F_{\mu\nu} F^{\mu\nu} + i Z_2\bar{\psi}\gamma^\mu\partial_\mu \psi + \sqrt{Z_1}Z_2 Z_e e \bar{\psi} \gamma^\mu A_\mu \psi
\end{equation}
$Z_1,Z_2$ and $Z_e$ are all fixed by renormalization to cancel the divergent parts of the loop integrals.
The ward identity says that $Z_2=\sqrt{Z_1}Z_2 Z_e$, or in other words
\begin{equation}
Z_e=\frac{1}{\sqrt{Z_1}}
\end{equation}
Since $Z_1$ as I have defined it leads to a residue $1/\sqrt{Z_1}$ in the photon propagator, this is equivalent to the equation you wrote above. (by the way this factor of $1/\sqrt{Z_1}$ will appear on all photon propagators, not just the on shell ones).
Note that as a consequence of the ward identity I can rewrite the last two terms in the lagrangian as
\begin{equation}
i Z_2\bar{\psi}\gamma^\mu\partial_\mu \psi + \sqrt{Z_1}Z_2 Z_e e \bar{\psi} \gamma^\mu A_\mu \psi= Z_2 i\bar{\psi}\gamma^\mu(\partial_\mu -i e A_\mu)\psi
\end{equation}
The right hand side is the gauge covariant derivative.
So when you adjust $Z_1$ to fix the norm of the photon's propagator to 1 (to match the LSZ formula and so forth), you also must adjust the electric charge by an appropriate amount. Alternatively you could go and compute the $eA\bar{e}$ three point function (using a regulator that preserves the gauge invariance, such as dim reg) and you would find that you had to renormalize it by this amount (which amounts to a of the ward identity at 1-loop).
Bonus comment: The constant $Z_1$ which appears in the lagrangian is a 'wave function renormalization', its just a rescaling of the field $A$ by $A\rightarrow \sqrt{Z_1} A$. How do we know what the right value for $Z_1$ is? It's a convention, and the convention is fixed by the LSZ formula. The LSZ formula tells you how to compute observables, and its based on a convention where the photon propagator has residue 1. So if there were no quantum corrections we would set $Z_1=1$. Loops correct the action, so we have to pick a value of $Z_1$ to cancel off the loop contributions. The total $Z$, $Z=Z_1+Z_{loops}$, will end up equaling 1, but we pick $Z_1$ to cancel out the loop contributions. However, we are using $Z_1$ in our definition of the free photon theory around which we are perturbing, and so we have to use $Z_1$ consistently every time we use a photon propagator. (There are actually many conventions for exactly where you put things, this is just one way to think of it.) However, worrying about putting factors of $Z_1$ on photon propagators (or choosing a convention where you put those factors somewhere else) only really starts mattering if you do higher loops, because $Z_1-1$ is already $O(\hbar)$. At your level the main point to realize is what's going on conceptually: the $Z_1$ in the action sets the size of ALL photon propagators (because its really the overall normalization of the photon field). We use the LSZ formula to fix the normalization, but that fixes the normalization for all propagators, not just the external ones.
Best Answer
The divergence is real but does not reflect a problem.
It arises because the potential has infinite range. It is similar to saying that in theory every snowball in the Andromeda galaxy is infinitesimally scattered by the sun, and it is just as meaningless because every really distant particle is more strongly perturbed by some other effect.
This divergence is not observed in interactions with a finite range such as the weak and effective (nuclear) strong interactions.