Taking the interior of a $0$-simplex to be the $0$-simplex itself is actually quite natural, if you think about it the following way:
The standard-$n$-simplex $\Delta^n$ is the subspace
$$
\textstyle
\Delta^n =
\{x=(x_0,\dots,x_n)\in\Bbb R^{n+1}\mid \sum_0^n x_i=1,\,x_i\ge0\,\forall i \}
$$
So a $\Delta^0$ is the point $\{1\}$ in $\Bbb R$, while the $\Delta^{-1}$ is the empty set $\emptyset$ (in $\Bbb R^0=\{0\}$). Since the boundary of an $n$-simplex is a union of $(n-1)$-simplices, it seems plausible that the boundary of a $0$-simplex should be empty.
Another approach is by looking at the interior of $\Delta^n$ relative to $\operatorname{Aff}(\Delta^n)$, the affine hull of the simplex, which is $\{(x_0,\dots,x_n)\mid \sum_0^n x_i = 1 \}$. This affine hull is topologically just $\Bbb R^n$, and seeing $\Delta^n$ as a subspace of the $n$-plane, its interior is
$$
\textstyle
\mathring{\Delta}^n = \{ (x_0,\dots,x_n) \mid \sum_0^n x_i=1, x_i>0\,\forall i \}
$$
Setting $n=0$, we get $\mathring\Delta^0=\{1\}$.
