Can $\mathbb{R}^{+}$ be divided into two disjoint sets so that each set is closed under both addition and multiplication?
This is an adaptation of the construction in the paper by Daniel Kane (linked in Jorge's answer).
Claim: There exists a non-trivial $\mathbb{Q}$-linear derivation on $\mathbb{R}$.
Before giving the proof, a non-trivial derivation $D$ lets us define a partition of $\mathbb{R}^+$ by $$ A=\big\{x\in \mathbb{R}^+:D(x) \geq 0\big\},\,\,\,\, B=\big\{x\in \mathbb{R}^+:D(x) < 0\big\}. $$ The sets $A$ and $B$ are closed under addition because $D$ is $\mathbb{Q}$-linear, and are closed under multiplication by the Leibniz rule. The partition is non-trivial because $D$ is non-trivial: if $x\in\mathbb{R}^+$ satisfies $D(x)\neq 0$, then $D(x)$ and $D(x^{-1})=-x^{-2}D(x)$ have opposite sign.
Proof of the claim: We construct a derivation with $D(\pi)=1$. Consider the set of pairs $(A,D)$, where $A$ is a subring of $\mathbb{R}$ containing $\pi$ and $D:A\to\mathbb{R}$ is a derivation satisfying $D(\pi)=1$. This set is non-empty because it contains $(\mathbb{Q}[\pi],\frac{d}{d\pi})$, and is partially ordered by extension. The set satisfies the hypotheses of Zorn's Lemma, so contains a maximal element $(A,D)$. It must be the case that $A=\mathbb{R}$, because if $x\in\mathbb{R}\backslash A$ we can extend $D$ to $A[x]$: if $x$ is transcendental over $A$ we set $D(x)=0$, and if $x$ is algebraic over $A$ with minimal polynomial $f$ we set $D(x)=-D(f)(x)/f'(x)$, where $D(f)$ is the polynomial obtained by applying $D$ to the coefficients of $f$.
This was answered three years ago at MO here. It is indeed possible.
The relevant links are this one, and this one, apparently a paper which characterizes the solutions is in preparation and will be published here, although its been in preparation for a long time now.
I'll leave my lame failed attempt:
Current attempt:
Think of $\mathbb R$ as a vector space over $\mathbb Q$, let $A$ and $B$ be the desired partition.
Clearly $span(A)\cup span(B)=\mathbb R$.
It follows wlog that $span(A)=\mathbb R$.
Pick a basis $a_i$ with index family $I$ of $\mathbb R$ consisting of elements of $A$.
It follows that every vector consisting of non-negative coefficients is in $A$.
Now suppose that a vector $a\in A$ contains a non-negative coefficient $\lambda_i$, then the vector $-a_i$ is also in $A$.
We can thus characterize $A$ as follows: Select a subset $X\subseteq I$, then $A$ is the set of vectors in $\mathbb R^+$ in which the coefficient belonging to each $a_i$ is non-negative for all $x\in X$.
We conclude that $B$ is the set of vectors in $\mathbb R^+$ such that at least one coefficient $a_x$ is negative. Of course if $|I|\neq 1$ we arrive at a contradiction, since we would be able to find two elements of $B$ whose sum is in $A$.
Conclusion: the sets $A$ and $B$ are formed in the following way:
Pick a basis of $\mathbb R$ with a distinguished vector $v$.
let $A$ be the subset of $\mathbb R^+$ containing non-negative $v$ coefficient.
let $B$ be the subset of $\mathbb R^+$ containing non-negative $v$ coefficient.