When in the propagator the interaction is considered, the interaction is formally characterized by the self-energy often called $\Sigma(p^2)$. It turns out that after Dyson resummation and mass correction a multiplicative term additional to the free (but with corrected mass) propagator appears, which is abreviated by $Z$ which is related to the self-energy by
$$Z = (1-\frac{d\Sigma(p^2)}{dp^2})^{-1}$$
This is indeed the same $Z$ than the one used to connect the Heisenberg field operator with the in-fields (or out-fields) which are treated as free fields so that we have asymptotically
$$\lim\limits_{t\rightarrow -\infty}\langle\gamma| \varphi(t) |\beta\rangle = \sqrt{Z} \langle \gamma | \varphi_{in}| \beta\rangle$$
where $\gamma$ and $\beta$ are variables which characterize Fock states.
The limit only makes sense using matrix elements of the Fock states, otherwise it would lead to contradictions.
The good news is that usually the self-energy and its derivative $\Sigma'(p^2)$ only depends logarithmically on the cut-off. Therefore, if the cut-off is chosen to be the Planck-mass $M_P\sim 10^{19}$GeV, the quantity $\Sigma'(p^2)$ is still small:
$$\delta_Z = \alpha^2 \log\frac{M_P}{M_{meas}}\quad \text{with}\quad Z=1-\delta_Z$$
where $\alpha$ is the coupling constant and $M_{meas}$ is the mass scale at which the computation is carried out (of course we assume that $\alpha \ll 1$ and higher order corrections $\sim \alpha^n$ can appear in the last equation too)
Then the requirement found in Bjorken& Drell (16.43)
$$0\le Z\lt 1$$
can be still fulfilled. At least this is my interpretation. Anyway, in all computations in QFT which deal with infinite integrals the only necessary requirement to make renormalization work, is that at the end of the computation the result is independent of the cut-off. A finite cut-off can be very well in place. In fact, in many textbooks on QFT a sentence can be found: After computation we can send the cut-off to infinity.
According to my interpretation this is not really necessary, anyway such a sentence does not change at all the computation before, for the computation to make sense the cut-off has to be finite. So it is okay. Tendentially we would like to have the cut-off at infinity, but according to Wilson approach to renormalization it can be conceptually useful to have a finite cut-off.
Choosing the cut-off at the Planck mass is sensible, because we expect the ultraviolet completion of QFT at Planck mass will solve the problem of the divergent integrals. How it is solved is still unknown, may be by string theory or by something else.
In condensed matter physics concepts like self-energy also appear, but the calculations always turn out to be finite, because in a condensed matter sample the smallest scale is at the atoms (~ 1 Å) so the cut-off is finite.
