The fundamental problem here is that many people, and also Pitts in his paper, are not careful about what theory they are currently talking about. "Quantization of Gauge Systems" by Henneaux and Teitelboim is actually careful about this, and their chapter 3 shows the correct resolution of this problem, even though Pitts cites it as an example for those who do not recognize the problem.
The claim "first-class constraints generate gauge transformations" is in the context of the extended action
$$ S_E[q^i, p_i, \lambda^i] = \int \left(p_i \dot{q}^i - H - \lambda^i \gamma_i\right)\mathrm{d}t,$$
where the $\gamma_i$ are the first-class constraints, $\lambda^i$ the Lagrange multipliers enforcing the constraints and for simplicity we assume there are no second-class constraints (as is the case in the example of electromagnetism). In this formulation, solutions to the equations of motion are tuples $(q(t), p(t), \lambda(t))$, and symmetries act on all these dynamical variables. The extended action is invariant under the infinitesimal local transformation $\delta_{\epsilon} F = \epsilon^i \{F,\gamma_i\}$ for $F(q,p)$ any function of the $q$ and $p$ only if one additionally lets the Lagrange multipliers transform as
$$ \delta_{\epsilon}\lambda^i = \dot{\epsilon}^i + {C^i}_{jk}\lambda^j\epsilon^k,$$
where $\{\gamma_i,\gamma_j\} = {C^k}_{ij}\gamma_k$. This set of symmetries reduces to the symmetries of the non-extended (canonical) action
$$ S_C[q^i,p_i,\bar{\lambda}^i] = \int \left(p_i \dot{q}^i - H - \lambda^{i'} \gamma_{i'}\right),$$
where the $i'$ index now only runs over the primary constraints, only after imposing the gauge conditions $\lambda^j = 0$ for all $j$ where $\gamma_j$ is not primary. It is the symmetries of the canonical action, not of the extended action, that directly translate to symmetries of the original Lagrangian action. The residual gauge symmetries of this action are those that preserve the conditions $\lambda^j = 0$ for the non-primaries, and are in general generated by a specific subset of combinations of the first-class constraints that other sources call the gauge generator(s).
For the concrete example Pitts is complaining about, the gauge transformations of the extended action of free electrodynamics are (see also H/T's chapter 19):
\begin{align}
\delta A_0 & = \epsilon^1 & \delta A_i & = \partial_i\epsilon^2 \\
\delta \lambda^1 & = \dot{\epsilon}^1& \delta\lambda^2 & = \dot{\epsilon}^2 - \epsilon^1
\end{align}
for two arbitrary functions of spacetime $\epsilon^1, \epsilon^2$. The preservation of the gauge condition $\lambda^2 = 0$ imposes $\epsilon^1 = \dot{\epsilon^2}$, so we are left with the residual gauge symmetry
$$ \delta A_\mu = \partial_\mu \epsilon^2 \quad \delta \lambda^1 = \ddot{\epsilon}^2,$$
which now has the familiar form for a gauge transformation of the 4-potential $A_\mu$ of Lagrangian electrodynamics.
The thing that seems to trip people up is that in the extended formalism, the quantity $E^i = F^{0i} = \partial^0 A^i - \partial^i A^0$, which they would like to identify as the electric field, is gauge-variant under transformations with $\epsilon^1 \neq \dot{\epsilon}^2$. How can the extended theory be "faithful" to our real, physical system if it turns gauge-invariant quantities into gauge-variant ones?
To see how, let us inspect the change of $E^i$ under an arbitrary transformation:
\begin{align}
\delta E_i & = \partial_0 \delta A_i - \partial_i \delta A_0 \\
& = \partial_i (\dot{\epsilon}^2 - \epsilon^1)
\end{align}
We note that this is precisely the spatial derivative of the transformation behaviour of $\lambda^2$, so this means that $E^i - \partial^i\lambda^2$ is a gauge-invariant observable, that turns into the residual gauge-invariant $E^i$ upon use of the gauge condition $\lambda^2 = 0$. So the extended theory does contain a gauge-invariant observable that is the electric field, it is just that its expression contains the auxiliary variable $\lambda^2$ which we have to eliminate to see whether or not this theory is equivalent to the Lagrangian theory that doesn't know about the $\lambda^i$.
