Suppose we can partition $\mathbb{R}$ into two subset $A,B$, both non empty and closed by sum and product. Let $0\in A$, and suppose that exists $b\in B$. Then $b^2\in B$. Now, $-b\in B$, cause if $-b\in A$ we would have $(-b)^2=b^2\in A$. Now, $0=b+(-b)\in B$, absurd. So $B$ must be empty, and the required partition doesn't exists.
Now we can ask the same question for $\mathbb{R}^+$. I guess that also in this case there are no such partition. Clearly the argument above doesn't work, cause we can't use negative numbers. Note also that a partition into subset closed only by sum exists: infact let $x_i$ an Hamel base of $\mathbb{R}$ over $\mathbb{Q}$: $$A=\{x\in\mathbb{R}^+:x=\alpha_1x_1+\ldots,\quad\alpha_1\geq 0\}$$ $$B=\{x\in\mathbb{R}^+:x=\alpha_1x_1+\ldots,\quad\alpha_1\leq 0\}$$ satisfies the request. Thank you for any suggestions.