As for your first question, one should define "standard", or acknowledge that a "canonical model" has been gradually established. As a comment indicated, it appears at least that the way you use IRWLS is rather standard.
As for your second question, "contraction mapping in probability" could be linked (however informally) to convergence of "recursive stochastic algorithms". From what I read, there is a huge literature on the subject mainly in Engineering. In Economics, we use a tiny bit of it, especially the seminal works of Lennart Ljung -the first paper was Ljung (1977)- which showed that the convergence (or not) of a recursive stochastic algorithm can be determined by the stability (or not) of a related ordinary differential equation.
(what follows has been re-worked after a fruitful discussion with the OP in the comments)
Convergence
I will use as reference Saber Elaydi "An Introduction to Difference Equations", 2005, 3d ed.
The analysis is conditional on some given data sample, so the $x's$ are treated as fixed.
The first-order condition for the minimization of the objective function, viewed as a recursive function in $m$,
$$m(k+1) = \sum_{i=1}^{N} v_i[m(k)] x_i, \;\; v_i[m(k)] \equiv \frac{w_i[m(k)]}{ \sum_{i=1}^{N} w_i[m(k)]} \qquad [1]$$
has a fixed point (the argmin of the objective function).
By Theorem 1.13 pp 27-28 of Elaydi, if the first derivative with respect to $m$ of the RHS of $[1]$, evaluated at the fixed point $m^*$, denote it $A'(m^*)$, is smaller than unity in absolute value, then $m^*$ is asymptotically stable (AS). More over by Theorem 4.3 p.179 we have that this also implies that the fixed point is uniformly AS (UAS).
"Asymptotically stable" means that for some range of values around the fixed point, a neighborhood $(m^* \pm \gamma)$, not necessarily small in size, the fixed point is attractive , and so if the algorithm gives values in this neighborhood, it will converge. The property being "uniform", means that the boundary of this neighborhood, and hence its size, is independent of the initial value of the algorithm. The fixed point becomes globally UAS, if $\gamma = \infty$.
So in our case, if we prove that
$$|A'(m^*)|\equiv \left|\sum_{i=1}^{N} \frac{\partial v_i(m^*)}{\partial m}x_i\right| <1 \qquad [2]$$
we have proven the UAS property, but without global convergence. Then we can either try to establish that the neighborhood of attraction is in fact the whole extended real numbers, or, that the specific starting value the OP uses as mentioned in the comments (and it is standard in IRLS methodology), i.e. the sample mean of the $x$'s, $\bar x$, always belongs to the neighborhood of attraction of the fixed point.
We calculate the derivative
$$\frac{\partial v_i(m^*)}{\partial m} = \frac {\frac{\partial w_i(m^*)}{\partial m}\sum_{i=1}^{N} w_i(m^*)-w_i(m^*)\sum_{i=1}^{N}\frac{\partial w_i(m^*)}{\partial m}}{\left(\sum_{i=1}^{N} w_i(m^*)\right)^2}$$
$$=\frac 1{\sum_{i=1}^{N} w_i(m^*)}\cdot\left[\frac{\partial w_i(m^*)}{\partial m}-v_i(m^*)\sum_{i=1}^{N}\frac{\partial w_i(m^*)}{\partial m}\right]$$
Then
$$A'(m^*) = \frac 1{\sum_{i=1}^{N} w_i(m^*)}\cdot\left[\sum_{i=1}^{N}\frac{\partial w_i(m^*)}{\partial m}x_i-\left(\sum_{i=1}^{N}\frac{\partial w_i(m^*)}{\partial m}\right)\sum_{i=1}^{N}v_i(m^*)x_i\right]$$
$$=\frac 1{\sum_{i=1}^{N} w_i(m^*)}\cdot\left[\sum_{i=1}^{N}\frac{\partial w_i(m^*)}{\partial m}x_i-\left(\sum_{i=1}^{N}\frac{\partial w_i(m^*)}{\partial m}\right)m^*\right]$$
and
$$|A'(m^*)| <1 \Rightarrow \left|\sum_{i=1}^{N}\frac{\partial w_i(m^*)}{\partial m}(x_i-m^*)\right| < \left|\sum_{i=1}^{N} w_i(m^*)\right| \qquad [3]$$
we have
$$\begin{align}\frac{\partial w_i(m^*)}{\partial m} = &\frac{-\rho''(|x_i-m^*|)\cdot \frac {x_i-m^*}{|x_i-m^*|}|x_i-m^*|+\frac {x_i-m^*}{|x_i-m^*|}\rho'(|x_i-m^*|)}{|x_i-m^*|^2} \\
\\
&=\frac {x_i-m^*}{|x_i-m^*|^3}\rho'(|x_i-m^*|) - \rho''(|x_i-m^*|)\cdot \frac {x_i-m^*}{|x_i-m^*|^2} \\
\\
&=\frac {x_i-m^*}{|x_i-m^*|^2}\cdot \left[\frac {\rho'(|x_i-m^*|)}{|x_i-m^*|}-\rho''(|x_i-m^*|)\right]\\
\\
&=\frac {x_i-m^*}{|x_i-m^*|^2}\cdot \left[w_i(m^*)-\rho''(|x_i-m^*|)\right]
\end{align}$$
Inserting this into $[3]$ we have
$$\left|\sum_{i=1}^{N}\frac {x_i-m^*}{|x_i-m^*|^2}\cdot \left[w_i(m^*)-\rho''(|x_i-m^*|)\right](x_i-m^*)\right| < \left|\sum_{i=1}^{N} w_i(m^*)\right|$$
$$\Rightarrow \left|\sum_{i=1}^{N}w_i(m^*)-\sum_{i=1}^{N}\rho''(|x_i-m^*|)\right| < \left|\sum_{i=1}^{N} w_i(m^*)\right| \qquad [4]$$
This is the condition that must be satisfied for the fixed point to be UAS. Since in our case the penalty function is convex, the sums involved are positive. So condition $[4]$ is equivalent to
$$\sum_{i=1}^{N}\rho''(|x_i-m^*|) < 2\sum_{i=1}^{N}w_i(m^*) \qquad [5]$$
If $\rho(|x_i-m|)$ is Hubert's loss function, then we have a quadratic ($q$) and a linear ($l$) branch,
$$\rho(|x_i-m|)=\cases{ (1/2)|x_i- m|^2 \qquad\;\;\;\; |x_i-m|\leq \delta \\
\\
\delta\big(|x_i-m|-\delta/2\big) \qquad |x_i-m|> \delta}$$
and
$$\rho'(|x_i-m|)=\cases{ |x_i- m| \qquad |x_i-m|\leq \delta \\
\\
\delta \qquad \qquad \;\;\;\; |x_i-m|> \delta}$$
$$\rho''(|x_i-m|)=\cases{ 1\qquad |x_i-m|\leq \delta \\
\\
0 \qquad |x_i-m|> \delta} $$
$$\cases{ w_{i,q}(m) =1\qquad \qquad \qquad |x_i-m|\leq \delta \\
\\
w_{i,l}(m) =\frac {\delta}{|x_i-m|} <1 \qquad |x_i-m|> \delta} $$
Since we do not know how many of the $|x_i-m^*|$'s place us in the quadratic branch and how many in the linear, we decompose condition $[5]$ as ($N_q + N_l = N$)
$$\sum_{i=1}^{N_q}\rho_q''+\sum_{i=1}^{N_l}\rho_l'' < 2\left[\sum_{i=1}^{N_q}w_{i,q} +\sum_{i=1}^{N_l}w_{i,l}\right]$$
$$\Rightarrow N_q + 0 < 2\left[N_q +\sum_{i=1}^{N_l}w_{i,l}\right] \Rightarrow 0 < N_q+2\sum_{i=1}^{N_l}w_{i,l}$$
which holds. So for the Huber loss function the fixed point of the algorithm is uniformly asymptotically stable, irrespective of the $x$'s. We note that the first derivative is smaller than unity in absolute value for any $m$, not just the fixed point.
What we should do now is either prove that the UAS property is also global, or that, if $m(0) = \bar x$ then $m(0)$ belongs to the neighborhood of attraction of $m^*$.
Best Answer
Yes, IRLS could be faster, as I said in my answer to your previous question. For example, if the log-likelihood is nearly quadratic (which it will usually be if you are able to start fairly close to the maximum and the sample size isn't very small), then it may converge in only a couple of steps. Note, in fact that on p240, Bishop says
Notice the "LS" part. IRLS proceeds by performing weighted least squares, but the weights to observations are updated each step (re-weighted).
The particular version of IRLS Bishop presents might possibly be Newton-Raphson, but this is not necessarily the case in general (it could be Fisher scoring for example, which is related to but slightly different from actual Newton Raphson). But yes, a single step of IRLS on a (already correctly-weighted) regression problem suffices.