Renormalization condition: why must be the residue of the propagator be 1?

The OS condition that $$ \frac{\partial\Sigma}{\partial p^2}|_{p^2=-m^2} = 0 $$ implies that the residue in the propagator remains equal to one.

Suppose that we used a different renormalization scheme in which our counter-terms contain no finite parts (e.g. MS scheme). In the OS scheme, we removed finite parts which were logarithmic in our artificial regularization scale $\mu$. In our new choice, the propagator might have a residue, say $R$.

This residue manifests itself in an irritating way; the field will be re-normalized such that $\phi = \frac{1}{\sqrt{R}} \phi_B$. In the LSZ formula, however, external lines contribute factors $R$ (from the KG equation cancelling the propagators). So external scalar lines contribute a factor $\sqrt{R}$ in the MS scheme.

So, whilst this choice in the OS scheme is somewhat arbitrary, it's convenient, because external scalar lines contribute a factor 1.

I'm trying to learn these points myself, so hopefully someone can expand/correct my answer where necessary...

The pole corresponds to an on-shell particle going from one point to another. Then, the residue effectively tells you how many of those particles are being transmitted. Since in your physical/renormalized theory, the propagator should correspond to $1$ quantum of the renormalized field being transmitted, you set the residue at the pole to $1$.