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I think it is not possible. Such a ring isomorphism $\phi$ should also preserve the order structure, because in both rings non-negative elements are exactlty the squares, and since both rings are divisible, it should also preserve the real scalar multiplication. The ring $S$ possesses all characteristic functions of singletons of $\mathbb{R}$. These $u:= \chi_{\{t\}}$ can be characterized in terms of the ordered $\mathbb{R}$-algebra structure, as e.g. those idempotents $u$ all of whose smaller positive elements are proportional to $u$ (that is: $0\le f\le u$ implies $f=\lambda u$). Since as we proved that $\phi$ is an ordered $\mathbb{R}$-algebras isomorphism, $\phi(u)$ has the same property of $u$, and since $X$ is Tychonov, $\phi(u)$ is also a characteristic function of a singleton of $X$. This way we may define a map $\psi:\mathbb{R}\to X$ such that for all $t\in\mathbb{R}$ one has $\phi(\chi_{\{t\}})= \chi_{\{\psi(t)\}}\in C(X)$, a characteristic function of an open point of $X$. So $C(X)$ also contains the characteristic function of $\psi(\mathbb{Q})$, and again by the order properties of $\phi$, this allows to conclude that $S$ contains the characteristic function of $\mathbb{Q}$, a contradiction.