After checking the arguments with @Jeff, I am willing to conclude that the issue in the question above is an extremely minor error in the original paper [BS91]. Both Jeff and I came up with "fixes" for the case of $u^{\rho}$ not continuous. Both are detailed below.
Update: I had the chance to speak to Professor Barles about this,
who was kind enough to suggest a way to improve my fix. The result is below.
Parsiad and Professor Barles' fix
Redefine the notion of an approximation scheme as a pair of functions
$$\underline{S},\overline{S}:(0,\infty)\times\overline{\Omega}\times\mathbb{R}\times B(\overline{\Omega})\rightarrow\mathbb{R}.$$
For fixed $\rho$, a solution of this approximation scheme is a bounded
function $u^{\rho}:\overline{\Omega}\rightarrow\mathbb{R}$ such that
$$
\underline{S}(\rho,x,(u^{\rho})^{*}(x),(u^{\rho})^{*})\leq0\leq \overline{S}(\rho,x,(u^{\rho})_{*}(x),(u^{\rho})_{*})\text{ for all }x\in\overline{\Omega}.
$$
Remark: Intuitively, $(u^{\rho})^*$ and $(u^{\rho})_*$ are simply upper and lower semicontinuous envelopes of the solution $u^{\rho}$ of an approximation scheme in the sense of [BS91].
Monotonicity is extended in the obvious way: an approximation scheme
$(\underline{S},\overline{S})$ is monotone if for each $S\in\{\underline{S},\overline{S}\}$,
\begin{gathered}
S(\rho,x,t,u)\leq S(\rho,x,t,v)\text{ if }u\geq v\\
\text{ for all }\rho>0\text{, }x\in\overline{\Omega}\text{, }t\in\mathbb{R}\text{, and }u,v\in B(\overline{\Omega}).
\end{gathered}
Stability is also extended in the obvious way: an approximation scheme $(\underline{S},\overline{S})$ is stable
if
\begin{gathered}
\text{for all }\rho>0\text{, there exists a solution }u^{\rho}\in B(\overline{\Omega})\text{,}\\
\text{with }u^{\rho}\text{ bounded independent of }\rho.
\end{gathered}
The consistency requirement is modified slightly. In particular, we
require that $\overline{S}$ (resp. $\underline{S}$) only satisfy (2.4a) (resp.
(2.4b)) of [BS91]. That is, an approximation scheme $(\underline{S},\overline{S})$
is consistent if for all $x\in\overline{\Omega}$ and $\varphi\in C_{b}^{\infty}(\overline{\Omega})$,
\begin{align*}
\limsup_{\substack{\rho\rightarrow0\\
y\rightarrow x\\
\xi\rightarrow0
}
}\frac{\overline{S}(\rho,y,\varphi(y)+\xi,\varphi+\xi)}{\rho} & \leq F^{*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x)\\
\text{and }\liminf_{\substack{\rho\rightarrow0\\
y\rightarrow x\\
\xi\rightarrow0
}
}\frac{\underline{S}(\rho,y,\varphi(y)+\xi,\varphi+\xi)}{\rho} & \geq F_{*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x).
\end{align*}
Proposition: Let $(\underline{S},\overline{S})$ be an approximation scheme that
is monotone, stable, and consistent (in the sense above). Assume $F=0$
satisfies a strong comparison principle (i.e., (2.5) of [BS91]).
Then, as $\rho\rightarrow0$, the solution $u^{\rho}$ of the scheme
converges locally uniformly to the unique viscosity solution of $F=0$ (in the sense of [BS91]).
Proof: With the setup above, the proof is essentially identical to that of [BS91] (and avoids the issue of being unable to take extrema).
Jeff's fix
Proposition: Theorem 2.1 of [BS91] holds if we replace (2.4) of [BS91] by the following modified consistency requirement
\begin{align*}
\limsup_{\substack{\rho\rightarrow0\\
y\rightarrow x\\
\xi\rightarrow0
}
}\frac{S(\rho,y,\varphi(y)+\xi+e^{-1/\rho},\varphi+\xi)}{\rho} & \leq F^{*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x) & \text{(2.4a}\text{)}\phantom{.}\\
\text{and }\liminf_{\substack{\rho\rightarrow0\\
y\rightarrow x\\
\xi\rightarrow0
}
}\frac{S(\rho,y,\varphi(y)+\xi-e^{-1/\rho},\varphi+\xi)}{\rho} & \geq F_{*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x) & \text{(2.4b}\text{)}.
\end{align*}
Proof: While the claim (2.7) cannot be proved, we can instead establish the existence of a sequence $(\rho_{n},y_{n})$ such that
$$
\rho_{n}\rightarrow0\text{, }y_{n}\rightarrow x_{0}\text{, and }(u^{\rho_{n}}-\varphi)(y_{n})+e^{-1/\rho_{n}}\geq(u^{\rho_{n}}-\varphi)(\cdot).
$$
To see this, let $(\rho_{n},x_{n})_{n}$ be a sequence with $\rho_{n}\rightarrow0$, $x_{n}\rightarrow x_{0}$, and $u^{\rho_{n}}(x_{n})\rightarrow\overline{u}(x_{0})$. Then, for each $n$, there exists $y_{n}$ such that
$$
(u^{\rho_{n}}-\varphi)(y_{n})+e^{-1/\rho_{n}}\geq\sup_{\overline{B(x_{0};r)}}\left\{ u^{\rho_{n}}-\varphi\right\} .
$$
Extract a subsequence of $(\rho_{n},x_{n},y_{n})_{n}$ converging
to some $(0,x_{0},\hat{y})$ and relabel it, with a slight abuse of
notation, $(\rho_{n},x_{n},y_{n})_{n}$. It follows that
\begin{multline*}
0=(\overline{u}-\varphi)(x_{0})=\lim_{n\rightarrow\infty}(u^{\rho_{n}}-\varphi)(x_{n})\leq\limsup_{n\rightarrow\infty}\left\{ (u^{\rho_{n}}-\varphi)(y_{n})+e^{-1/\rho_{n}}\right\} \\
\leq\limsup_{\substack{\rho\rightarrow0\\
y\rightarrow\hat{y}
}
}\left\{ (u^{\rho}-\varphi)(y)+e^{-1/\rho}\right\} =\limsup_{\substack{\rho\rightarrow0\\
y\rightarrow\hat{y}
}
}(u^{\rho}-\varphi)(y)=(\overline{u}-\varphi)(\hat{y}).
\end{multline*}
Because it was assumed that $x_{0}$ is a strict local maximum of
$\overline{u}-\varphi$, we get $\hat{y}=x_{0}$, as desired.
Now, the remainder of the proof in [BS91] is modified as follows. Let $\xi_{n}=(u^{\rho_{n}}-\varphi)(y_{n})+e^{-1/\rho_{n}}$.
Then, $u^{\rho_{n}}\leq\varphi+\xi_{n}$ and hence
$$
0=S(\rho_{n},y_{n},u^{\rho_{n}}(y_{n}),u^{\rho_{n}})\geq S(\rho_{n},y_{n},\varphi(y_{n})+\xi_{n}-e^{-1/\rho_{n}},\varphi+\xi_{n}).
$$
The desired result now follows by dividing by $\rho$, taking limit inferiors, and applying the consistency condition. $\square$
Remark: the exponential decay is chosen as it does not cause any problems in
creating approximation schemes, practically speaking. For example, consider the approximation $$\frac{u^\rho(x+\rho) - u^\rho(x)}{\rho}.$$ If we substitute the corresponding entities in the revised consistency requirement, we get $$\frac{\varphi(x+\rho) - \varphi(x) \pm e^{-1/\rho}}{\rho} = \varphi^\prime(x) + \mathcal{O}(\rho)$$ by a Taylor approximation.
