In the theory of free scalar bosons (KG field) it is a basic result that the propagator $\Delta(p)$ has poles at $p^2=m^2$, with residue $1$ (or any other constant, depending on conventions). Thinking of $p^2$ as a variable is acceptable because it can be shown that the propagator is a function of $p^2$, even in the interacting case.
On the other hand, in the theory of free spinor fields (Dirac field), we can show that the propagator is not a function of $p^2$, but of $\not p$, so we can't use the $p^2$ trick. Conventions aside, the propagator is $$ S(p)=\frac{\not p+m}{p^2-m^2+i \epsilon} $$
It is clear that we cannot think of poles as $p^2\to m^2$, because $S(p)$ depends on $p$ in a non-so-trivial way. The usual solution is to think of $S$ as a function of $\not p$, as if $\not p$ was a number instead of a matrix, thus writing $$ S(\not p)=\frac{1}{\not p-m+i\epsilon} $$ and, again, we find a pole at $\not p=m$, with residue $1$. I think this is not mathematics. This is just nonsensical to me (may be I'm too skeptic, and we can give a meaning to this last formula).
In an interacting theory, we define the field strength normalisation $Z$ as the residue of the propagator at the poles: $$ \Delta(p^2)=\frac{Z}{p^2-m^2+i\epsilon}+\int_{M_\text{thresh}^2}^\infty \mathrm d\mu^2\rho(\mu^2)\frac{1}{p^2-\mu^2+i\epsilon} $$ $$ S(\not p)=\frac{Z}{\not p-m+i\epsilon}+\int_{M_\text{thresh}^2}^\infty \mathrm d\mu^2\frac{\not p\rho_1(\mu^2)+\mu \rho_2(\mu^2)}{\not p^2-\mu^2+i\epsilon} $$
Now, I don't understand what's the proper definition of $Z$ as a residue. In what sense is it a residue? What's the variable? I just can't accept it is a residue as $\not p\to m$. Do we really take this "think of $\not p$ as a variable'' seriously? Is it possible to formalise these matters, by thinking of a residue for an actual variable (as $p_0$)?
Perhaps the $\not p$ trick is just that: a trick, which may simplify calculations, but such that it can be shown that the actual result, found by a more standard procedure, is the same. Is this the case? If so, what is the correct procedure?
(I would really appreciate if the answers don't assume that we can boost to the rest frame of the particle, as I'd like it to be as general as posible. I want to take into account the possibility of $m=0$, so please don't boost into $k=(m,\boldsymbol 0)$ if it's not really necessary)
Answer
Very good question, OP! Your scepticism is most certainly justified. The good news is, someone already has addressed your concerns. You can find the answer in Ticciati's Quantum Field Theory for Mathematicians, section 10.13.
Long story short: given \begin{equation} S(\not p)=\frac{1}{\not p-m-\Sigma(\not p)+i\epsilon} \end{equation} you can always parametrise the matrix $\Sigma$ as \begin{equation} \Sigma(\not p)=a(p^2)1+b(p^2)\not p \end{equation} for a pair of scalar functions $a,b$. This general expression is the result of Lorentz- and parity-invariance. But you already know that.
With this, you can rationalise the denominator into \begin{equation} S(\not p)=\frac{i(\not p+\alpha)}{(1-b)(p^2-\alpha^2)+i\epsilon} \end{equation} where \begin{equation} \alpha(p^2)\overset{\mathrm{def}}=\frac{m+a(p^2)}{1-b(p^2)} \end{equation}
The required pole at $p^2=m^2$ implies \begin{equation} {\color{red}{\alpha(m^2)=m}},\tag1 \end{equation} that is, \begin{equation} S(\not p)=\frac{1}{(1-b)(1-2m\alpha')}\frac{i(\not p+m)}{p^2-m^2+i\epsilon}+\mathcal O(1) \end{equation}
If the residue is to be equal to that of a normalised field, that is, \begin{equation} S(\not p)=\frac{i(\not p+m)}{p^2-m^2+i\epsilon}+\mathcal O(1) \end{equation} then we must have \begin{equation} {\color{red}{(1-b(m^2))(1-2m\alpha'(m^2))=1}}\tag2 \end{equation}
So far so good: we have rigorously characterised the normalisation conditions of a bispinor field.
The key point is the following: if we introduce the formal complex variable $\not p\in\mathbb C$, we can formally write $(1),(2)$ as \begin{equation} \begin{aligned} \Sigma(m)&=0\\ \Sigma'(m)&=0 \end{aligned} \end{equation} as can be checked by a straightforward computation. This justifies the introduction of the formal complex variable $\not p$: the formal manipulations of this variable are equivalent to the more correct procedure of introducing the pair of scalar functions $a,b$, and the end result is the same. All's well that ends well I guess.
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