Is a local redefinition $q\to-q$ of "charge" $ \to$ "minus charge" a gauge transformation? This is claimed in the Scientific American article "Q&A: Lawrence Krauss on The Greatest Story Ever Told" published on March 21, 2017.

It says, “I could change the sign of each electric charge in nature locally. But I have to have a rule book.” What's the rule book? In this case, it’s the electromagnetic field. 

However looking at explicit $U(1)$ gauge transformations
$$\psi \to e^{ia(x)} \psi ,$$
$$ A\to A+ \partial_\mu a(x). $$
I fail to see how such an interpretation is possible
 A: That really depends on what you mean by ''Gauge Transformation''.
I will shortly give the mathematical description, which is also the one
used in (almost any) physics-book (but in a less technical setting) and once we have this, it's clear that
''charge'' $\rightarrow$ -''charge'' cannot be a gauge transformation.
For the general setting, let us model electromagnetism not on Minkowski
Space $\mathbb{R}^{1,3}$ but on $\mathbb{R}^{1,3} \times \mathrm{U}(1)$.
(In the following I will use a little bit of gauge theory, but try to minimize the technical details as best as possible.)
Since $\psi$ does transform under a gauge transformation, which are a change 
of phase, it doesn't seem too off to view $\psi$ as a function which depends
on $(x_{\mu},e^{i\theta})$. The latter parameter just explicilty encodes the
phase at each point in spacetime.
Thus, if you change the phase, which is what we will call a gauge transformation we have to specify the way in which the field changes.
The general setting would look like this:
$$ \psi (x_{\mu},e^{i \theta} \cdot e^{i\phi(x_{\mu})}) = \Lambda(e^{i \phi (x_{\mu})})\psi (x_{\mu}, e^{i \theta})$$
Where $\Lambda$ is a representation of $\mathrm{U}(1)$ acting on the vector
space in which $\psi$ takes it's values (pre second quantisation).
Now let us assume the representation $\Lambda$ be irreducible and one
dimensional (by which I mean, it is given by $\Lambda (e^{i \phi (x_{\mu})})\psi = e^{ \lambda (i \phi (x_{\mu}))} \cdot \psi
 $ for some function $\lambda$.
Then $\lambda$ is given by multiplication
with an Integer $e \in \mathbb{Z}$, giving
$$ \psi (x_{\mu},e^{i \theta} \cdot e^{i\phi(x_{\mu})}) = e^{i \cdot e \cdot \phi (x_{\mu})}\psi (x_{\mu}, e^{i \theta}).$$
Finally, $e$ is the elementary (electromagnetic) charge of the particle described by $\psi$.
Thus, changing the charge would be a change of representation and not a 
gauge transformation, since gauge transformations do not touch the representation, hence, they don't touch the charge.
(going from $e$ to $-e$ would be done by passing from $\Lambda$
to $\Lambda$* which is the complex-conjugate representation)
I'd like to add, that the change of representation described above
would correspond to a change of particle, for example passing from the 
$\pi^+$ to the $\pi^-$-meson.
