The range is twice the maximum absolute deviation from the mid-range, a tenable 'central value'; so on the face of it ought to pass muster as a measure of dispersion around that.
Formal criteria that measures of location & dispersion ought to meet have been proposed in papers by Bickel & Lehmann—see
my answers to Is mode a measure of central tendency? & Is it really appropriate to say "standard deviation" is variation or dispersion. The mid-range & range meet these respective criteria.†
Note B. & L. appear to side with you in defining dispersion around a measure of location (they distinguish it from spread throughout a distribution). As @NickCox points out, a location measure may not appear explicitly in a particular definition of a dispersion measure; what matters is that the ordering of distributions by their range, say, can't contradict the partial ordering based on probabilities of absolute deviations from the mid-range, whereas it can contradict that based on probabilities of absolute deviations from the mean.
† The mid-range $\mu(X)$ is shift equivariant,
$$\mu(X + b) = \frac{\min(X+b) + \max(X+b)}{2}= \frac{(\min(X)+b) + (\max(X) + b)}{2} = \mu(X) + b$$
scale equivariant,
$$\mu(aX) = \frac{\min(aX) + \max(aX)}{2} = \frac{a\min(X) + a\max(X)}{2} =a\mu (X)\;\forall a > 0$$
& reflection equivariant
$$\mu(-X) = \frac{\min(-X) + \max(-X)}{2} = \frac{-\max(X) - \min(X)}{2} =-\mu (X)$$
Stochastic dominance of $X$ over $Y$ implies the mid-range of $X$ is no less than that of $Y$:
$$\begin{align}
\mu(X) -\mu(Y) &= \frac{\min(X)+\max(X)}{2} - \frac{\max(Y)+\min(Y)}{2}\\ &=\frac{\min(X)-\min(Y)}{2} + \frac{\max(X)-\max(Y)}{2}
\end{align}
$$
but for $X$ to dominate $Y$, both $\min(X)\geq\min(Y)$ & $\max(X)\geq\max(Y)$; therefore $\mu(X) \geq\mu(Y)$
The range $\sigma$ is shift-invariant,
$$\sigma(X + b) = \max(X+b) - \min(X+b) = (\max(X) + b) - (\min(X) +b)= \sigma (X)$$
scale-equivariant,
$$\sigma(aX) = \max(aX) - \min(aX) = a\max(X) - a\min(X) = a\sigma (X)\;\forall a > 0$$
& reflection-invariant:
$$\sigma(-X) = \max(-X) - \min(-X) = -\min(X) + \max(X)= \sigma (X)$$
Stochastic dominance of the absolute deviation of $X$ from the mid-range of $X$ over the absolute deviation of $Y$ from the mid-range of $Y$ implies the range of $X$ is no less than that of $Y$:
$$\begin{align}
\sigma(X) - \sigma(Y) &= \max(X) - \min(X) - (\max(Y)-\min(Y))\\
&= \max(X) - (2\mu(X) - \max(X)) - (\max(Y) - (2\mu(Y) - \max(Y)))\\
&= 2(\max(X) - \mu(X) - (\max(Y)-\mu(Y)))
\end{align}
$$
But for $|X-\mu(X)|$ to dominate $|Y-\mu(Y)|$, $\max(X) - \mu(X) \geq \max(Y)-\mu(Y)$; therefore $\sigma(X) \geq \sigma(Y)$