$\newcommand{\om}{\omega} \newcommand{\Om}{\Omega} \newcommand{\N}{\mathbb N} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\R}{\mathbb R}$ The latest clarification by the OP appears useful, giving rise to the following construction.

Define the class $RV$ as follows.

Let $\Om:=\{0,1\}^\N$, let $F$ be the Borel $\si$-algebra with respect to the product topology over $\Om$, and let $P$ be the product probability measure $\la^{\otimes\N}$, where $\la$ is the uniform distribution on $\{0,1\}$. Clearly, the probability space $(\Om,F,P)$ is isomorphic to the Lebesgue probability space over the interval $[0,1]$.

Say that a subset $S$ of $\N$ is thin if the cardinality of $S\cap[n]$ is $o(n)$ as $n\to\infty$, where $[n]:=\{1,\dots,n\}$.

Let now $RV$ be the set of all (say real-valued) random variables (r.v.'s) $A$ defined on the probability space $(\Om,F,P)$ such that for some thin $S=S_A\subset\N$, some Borel function $f=f_A\colon\{0,1\}^S\to\R$, and all $\om\in\Om$ we have $$A(\om)=f(\om|_S);$$ that is, $A\in RV$ iff $A(\om)$ depends only on the values of the function $\om$ on a thin subset $S$ of $\N$.

Clearly, for any $k\in\N$, any r.v.'s $A_1,\dots,A_k$ in $RV$, and any Borel function $g\colon\R^k\to\R$, we have $g(A_1,\dots,A_k)\in RV$. This follows because the union of finitely many thin subsets of $\N$ is thin.

Moreover, for any $k\in\N$ and any probability distribution $\nu$ on $\R^k$, there are r.v.'s $A_1,\dots,A_k$ in $RV$ such that the "joint" distribution of $(A_1,\dots,A_k)$ is $\nu$. This follows because there are infinite thin subsets of $\N$.

Furthermore, for any r.v.'s $A$ and $B$ in $RV$ there is a r.v. $K\in RV$ such that $K$ is independent of $(A,B)$ and $P(K=1)=P(K=2)=P(K=3)=1/3$. Letting then $$C:=A\,1(K=1)+B\,1(K=2)+(A+B)\,1(K=3),$$ we get a r.v. $C\in RV$ such that "$C$ is $A$ with probability $1/3$, $C$ is $B$ with probability $1/3$, and $C$ is $A+B$ with probability $1/3$", as desired.

In view of the Borel isomorphism theorem, here instead of real-valued r.v.'s we may consider r.v.'s with values in arbitrary Polish spaces.

$\newcommand{\om}{\omega} \newcommand{\Om}{\Omega} \newcommand{\N}{\mathbb N} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\R}{\mathbb R}$ The latest clarification by the OP appears useful, giving rise to the following construction.

Define the class $RV$ as follows.

Let $\Om:=\{0,1\}^\N$, let $F$ be the Borel $\si$-algebra with respect to the product topology over $\Om$, and let $P$ be the product probability measure $\la^{\otimes\N}$, where $\la$ is the uniform distribution on $\{0,1\}$. Clearly, the probability space $(\Om,F,P)$ is isomorphic to the Lebesgue probability space over the interval $[0,1]$.

Say that a subset $S$ of $\N$ is thin if the cardinality of $S\cap[n]$ is $o(n)$ as $n\to\infty$, where $[n]:=\{1,\dots,n\}$.

Let now $RV$ be the set of all (say real-valued) random variables (r.v.'s) $A$ defined on the probability space $(\Om,F,P)$ such that for some thin $S=S_A\subset\N$, some Borel function $f=f_A\colon\{0,1\}^S\to\R$, and all $\om\in\Om$ we have $$A(\om)=f(\om|_S);$$ that is, $A\in RV$ iff $A(\om)$ depends only on the values of the function $\om$ on a thin subset $S$ of $\N$.

Clearly, for any $k\in\N$, any r.v.'s $A_1,\dots,A_k$ in $RV$, and any Borel function $g\colon\R^k\to\R$, we have $g(A_1,\dots,A_k)\in RV$. This follows because the union of finitely many thin subsets of $\N$ is thin.

Moreover, for any $k\in\N$ and any probability distribution $\nu$ on $\R^k$, there are r.v.'s $A_1,\dots,A_k$ in $RV$ such that the "joint" distribution of $(A_1,\dots,A_k)$ is $\nu$. This follows because there are infinite thin subsets of $\N$.

Further, for any countable set $T$ and consistent family of finite-dimensional probability distributions on $\R^S$ indexed by finite subsets $S$ of $T$, there is a family $(A_t)_{t\in T}$ of r.v.'s in $RV$ with the given finite-dimensional distributions. This follows because there is a countable set of disjoint infinite thin subsets of $\N$.

Furthermore, for any r.v.'s $A$ and $B$ in $RV$ there is a r.v. $K\in RV$ such that $K$ is independent of $(A,B)$ and $P(K=1)=P(K=2)=P(K=3)=1/3$. Letting then $$C:=A\,1(K=1)+B\,1(K=2)+(A+B)\,1(K=3),$$ we get a r.v. $C\in RV$ such that "$C$ is $A$ with probability $1/3$, $C$ is $B$ with probability $1/3$, and $C$ is $A+B$ with probability $1/3$", as desired.

In view of the Borel isomorphism theorem, here instead of real-valued r.v.'s we may consider r.v.'s with values in arbitrary Polish spaces.