For every positive integer $m$, we know that $m$ and $S(m)$ have the same remainder $r$ modulo $3$. Then, $m + 2S(m)$ has the same remainder as $3r$ and so, it is a multiple of $3$. Therefore, if a number $n$ is rioplatense, then it is a multiple of $3$.

Now we are going to show that every multiple of $3$ is rioplatense. We will prove by induction on $k$ that every integer $n$ divisible by $3$ with at most $k$ digits can be written as $n = m + 2S(m)$ for an integer $m$ with at most $k$ digits.

For $k=1$, the result is immediate, since $3 = 1 + 2S(1)$, $6 = 2 + 2S(2)$ and $9 = 3 + 2S(3)$. Assume the result holds for $k \ge 1$ and consider an integer $n$ multiple of $3$ with $k+1$ digits. We have $10^k + 2 \le n < 10^{k+1}$, since $10^k + 2 = 10\cdots02$ is the smallest multiple of $3$ with $k+1$ digits. Let us call $N = 10^k + 2$. When dividing $n$ by $N$, we obtain a quotient $q$ and a remainder $r$. Note that $1 \le q \le 9$, since $n < 10^{k+1} < 10N$. On the other hand, $r < 10^k + 2$ and it is a multiple of $3$, since $r = n - Nq$ and both $n$ and $N$ are multiples of $3$, which implies that $r \le 10^k - 1$ and so, it has at most $k$ digits. By the induction assumption, there exists $t = \overline{a_{k-1} \cdots a_1 a_0}$ such that $r = t + 2S(t)$.

Take $m = \overline{a_k a_{k-1} \cdots a_1 a_0}$, where $a_k$ is the quotient $q$. Note that $m$ has $k+1$ digits and
$$\begin{align*} m + 2S(m) &= \overline{a_k a_{k-1} \cdots a_1 a_0} + 2(a_k + a_{k-1} + \cdots + a_1 + a_0) = \\ &= 10^k a_k + \overline{a_{k-1} \cdots a_1 a_0} + 2a_k + 2(a_{k-1} + \cdots + a_1 + a_0) = \\ &= (10^k + 2)a_k + t + 2S(t) = Nq + r = n, \end{align*}$$
as desired.