Yes, it is possible to construct such a partition. See this note by Daniel M. Kane. Any proof would have to involve the axiom of choice, otherwise you can't partition the positive reals into subsets closed under addition, let alone both addition and multiplication.
Added later: There is a recent paper, "Decomposing the real line into Borel sets closed under addition", where the authors show that the only partitions of $\mathbb R$ into countably many (Borel) sets closed under addition are of the form $\mathbb R_+\cup \{0\}\cup \mathbb R _-$, etc. And they mention that the structure of all partitions of $\mathbb R_+$ into two sets closed under addition and multiplication was determined in a paper by G. Kiss, G. Somlai, T. Terpai, which is still in preparation. (Here is G. Kiss's website.)