A "classical" view
Although it's not really meaningful to take the "classical" limit of the weak force, we can make an attempt to view it classically by studying something physicists call the Klein-Gordon equation, which describes the propagation of a massive particle. In our universe, for the case of a time-independent potential, it takes the form
$$\nabla^2\phi=\frac{m^2c^2}{\hbar^2}\phi\tag{1}$$
where $m$ is the mass of the particle. This yields the solution
$$\phi_Y(r)\sim\frac{1}{r}e^{-\alpha mr}$$
where $\alpha=c/\hbar$; we refer to $\phi_Y$ as a Yukawa potential (see these notes for a derivation of how it can be derived from the Klein-Gordon equation).
If we try to formulate the Klein-Gordon equation in Egan's universe, we get
$$\nabla^2\phi=-\frac{m^2c^2}{\hbar^2}\phi\tag{2}$$
which is a special case of what Egan refers to as the "Riemann scalar wave equation" if we set $\omega_m\equiv\frac{mc}{\hbar}$, remove the time dependence and set the current to $\mathbf{j}=0$. Solving this gives us
$$\phi_E(r)\sim\frac{\cos(\omega_mr)}{r}$$
which is Egan's expression for his version of the Coulomb potential. (Note that in both cases, if we set $m=0$, we recover the Coulomb potential for a massless photon.)
Effectively, if you try to view the weak force classically, you find that it ends up acting essentially the same as the electromagnetic force. This makes sense; in both cases, we're dealing with forces mediated by massive spin-1 gauge bosons. From this perspective, it at first looks like the weak force could indeed have infinite range.
The Lagrangian
Egan apparently goes into some detail on the quantum mechanical Lagrangian describing the fields in his universe. If we take out the electromagnetic part, we get
$$\mathcal{L}_{\text{EM}}=\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{1}{2}m_{\text{ph}}^2A_{\mu}A^{\mu}\tag{3}$$
with $m_{\text{ph}}$ the mass of the photon. To me, this looks exactly like what we get if we try to naïvely add a massive photon to a theory of electromagnetism in our universe - the Proca action - multiplied by $-1$. Now, you can show by picking a certain gauge transformation$^{\dagger}$ that this Lagrangian doesn't have the $\text{U}(1)$ gauge symmetry associated with the electromagnetic force in our universe; rather, it has the symmetry associated with the group $\mathbf{R}$ under addition (perhaps it's not surprising, as the two are in a sense quite similar).
I bring up $\mathcal{L}_{\text{EM}}$ to make the point that the Lagrangian describing the weak force may not look much like the weak force Lagrangian folks are used to in our universe insofar as it would not have to obey the same symmetry as ours does ($\text{SU}(2)$), which is why our universe requires the Higgs. Equation $(3)$ is a perfectly valid Lagrangian depending on whether you care about certain properties of your theory. This makes it hard to talk about what Egan's weak force Lagrangian would look like. In other words, looking at the Lagrangian tells us very little, or, to be more honest, nothing. We leave with less insight than we started with.
A quantum view
Egan doesn't seem to elaborate on interactions between particles in his universe, nor the masses of particular elementary particles (aside from photons). This is unfortunate, because those are the key things we would need to know to get the full picture of the weak force - the classical picture I began with is incomplete at best, and misleading at worst. What we can say is that it seems unlikely that there is anything in his universe that changes how particles can interact and decay, and these are what truly limit the range of the weak force.
Assuming that the decay widths $\Gamma$ are the same for the W and Z bosons as they are in our universe, the particles would still have lifetimes of $\tau\sim\hbar/\Gamma$. They would still be quite massive and would still decay quickly, traveling finite - and small - distances before decaying. Nothing in any of Egan's pages indicates otherwise, meaning that the weak force is, we can assume, still an extremely short-range force, no matter what any version of the Klein-Gordon equation says. Classical assumptions and results can only get you so far.
$^{\dagger}$We can pick the transformation $A_{\mu}\to A_{\mu}-\partial_{\mu}\eta$ for any $\eta$, and expand out $F_{\mu\nu}$ as
$$F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$$
and see that $\mathcal{L}_{\text{EM}}$ does not retain its original form if we have the mass term - it's algebra after this.
