Here's a potentially trivial answer.

Notations

Let $X = \mathbb{R}^{[0,1]}$ be the space of all real valued functions. Let $S\subsetneq X$ be the subspace of step functions. Elements of $X$ may be partially ordered by the product order: $$f \preceq g \iff \forall z\in[0,1] (f(z) \leq g(z)).$$ Given $f,g\in X$ denote by $[f,g]$ the closed order interval $$ [f,g] := \{h\in X | f\preceq h \preceq g\}. $$ Denote by $I_S$ the set of all (non-empty) intervals with both end points in $S$.

We can define a width functional $w:I_S \to [0,\infty)$ by $$ w([s_1, s_2]) := \int s_2 - s_1 ~dx $$ This is well-defined as $s_2 - s_1$ is a step function.

Pseudo-metric

Consider the function $\rho:X^2 \to [0,\infty]$ defined by $$ \rho(f,g) = \begin{cases} 0 & f = g \\ \inf \{ w(i) | i\in I_S; f\in i; g\in i\} & f\neq g\end{cases} $$ We take the convention that the infimum of the empty set is $+\infty$.

This function is trivially symmetric and non-negative. It vanishes on the diagonal by definition. And triangle inequality is satisfied by the observation that if $[s_1, s_2] \cap [t_1,t_2] \neq \emptyset$, then at every point $x\in [0,1]$, $$ \max(s_2(x), t_2(x)) - \min(s_1(x), t_1(x)) \leq s_2(x) - s_1(x) + t_2(x) - t_1(x) $$

So $\rho$ defines a pseudo-metric on $X$.

Closure of $S$ under $\rho$

The closure of $S$ under $\rho$ is simply the set of all points $f\in X$ such that $\inf_{s\in S} \rho(s,f) = 0$. This is in particular satisfied if for every $\epsilon > 0$ there exists $s_1 \preceq f \preceq s_2$ such that $w([s_1,s_2]) < \epsilon$. This is precisely the statement that $f$ is Darboux integrable.

Extension of Riemann integral

The Riemann integral is uniformly continuous on $S$, pretty much by definition. Let $s_1 \neq s_2\in S$, the smallest interval in $I_S$ that contains both $s_1$ and $s_2$ is $[s_1\wedge s_2, s_1\vee s_2]$. Its width is $\int |s_2 - s_1| ~dx$.

In other words, restricted to $S$, we have that $\rho$ is the $L^1$ distance between step functions. And so the Riemann integral is Lipschitz continuous with respect to the pseudometric $\rho$, and hence uniformly continuous with respect to it, and hence has unique continuous extension to the closure of $S$.

Here's a potentially trivial answer.

Notations

Let $X = \mathbb{R}^{[0,1]}$ be the space of all real valued functions. Let $S\subsetneq X$ be the subspace of step functions. Elements of $X$ may be partially ordered by the product order: $$f \preceq g \iff \forall z\in[0,1] (f(z) \leq g(z)).$$ Given $f,g\in X$ denote by $[f,g]$ the closed order interval $$ [f,g] := \{h\in X | f\preceq h \preceq g\}. $$ Denote by $I_S$ the set of all (non-empty) intervals with both end points in $S$.

We can define a width functional $w:I_S \to [0,\infty)$ by $$ w([s_1, s_2]) := \int s_2 - s_1 ~dx $$ This is well-defined as $s_2 - s_1$ is a step function.

Pseudo-metric

Consider the function $\rho:X^2 \to [0,\infty]$ defined by $$ \rho(f,g) = \begin{cases} 0 & f = g \\ \inf \{ w(i) | i\in I_S; f\in i; g\in i\} & f\neq g\end{cases} $$ We take the convention that the infimum of the empty set is $+\infty$.

This function is trivially symmetric and non-negative. It vanishes on the diagonal by definition. And triangle inequality is satisfied by the observation that if $[s_1, s_2] \cap [t_1,t_2] \neq \emptyset$, then at every point $x\in [0,1]$, $$ \max(s_2(x), t_2(x)) - \min(s_1(x), t_1(x)) \leq s_2(x) - s_1(x) + t_2(x) - t_1(x) $$

So $\rho$ defines a pseudo-metric on $X$.

Closure of $S$ under $\rho$

The closure of $S$ under $\rho$ is simply the set of all points $f\in X$ such that $\inf_{s\in S} \rho(s,f) = 0$. This is in particular satisfied if for every $\epsilon > 0$ there exists $s_1 \preceq f \preceq s_2$ such that $w([s_1,s_2]) < \epsilon$. This is precisely the statement that $f$ is Darboux integrable.

Extension of Riemann integral

The Riemann integral is uniformly continuous on $S$, pretty much by definition. Let $s_1 \neq s_2\in S$, the smallest interval in $I_S$ that contains both $s_1$ and $s_2$ is $[s_1\wedge s_2, s_1\vee s_2]$. Its width is $\int |s_2 - s_1| ~dx$.

In other words, restricted to $S$, we have that $\rho$ is the $L^1$ distance between step functions. And so the Riemann integral is Lipschitz continuous with respect to the pseudometric $\rho$, and hence uniformly continuous with respect to it, and hence has unique continuous extension to the closure of $S$.

Final Remarks

A key point is that the pseudometric $\rho$ is not translation invariant in $X$. Let $f = 0$ and $g$ be the characteristic function of $x = 1/2$, then $\rho(f,g) = 0$. But $\rho(f + D, g+D) = 1$ where $D$ is the Dirichlet function.

Indeed, this is why this answer is sort of "fake". The pseudometric is designed so that every non-Riemann integrable function is an isolated point: if $f$ is not Riemann integrable, then there exists some $\epsilon > 0$ such that for every pair of step functions $s_1 \preceq f \preceq s_2$ we have that $\int s_2 - s_1 ~dx \geq \epsilon$. And therefore for any $g \neq f$, we have that $\rho(f,g) \geq \epsilon$.