Given a topological ring $R$ and $U,V$ open subsets, we can show that $U+V$ is an open subset due to the fact that $x\mapsto x+y$ is a homeomorphism for every $y \in R$. Since, in general, $R$ is not a division ring, it is not easy whether $UV$ is open or not.
My intuition tells me there must exist $R$ and $U,V$ open subsets such that $UV$ is not open, but I can't construct it.
Here is a silly example: Let $A$ be any non-discrete topological abelian group. Endow it with the zero multiplication. Then $A$ is a non-unital topological ring and $UV=\{0\}$ is not open for all non-empty subsets $U,V$.
In order to get a unital example, consider the unitalization $\widetilde{A}$. The underlying abelian group is just $\mathbb{Z} \times A$, the multiplication is $(n,a) (m,b) = (nm,nb+ma)$, so that the unit is $(1,0)$. Let's endow $\mathbb{Z} \times A$ with the product topology, where $\mathbb{Z}$ carries the discrete topology. Then $\widetilde{A}$ is a unital topological ring. If $U,V$ are any non-empty open subsets of $A$, then $\{0\} \times U $ and $\{0\} \times V $ are open subsets of $\widetilde{A}$ with $(\{0\} \times U) \cdot (\{0\} \times V) = \{(0,0)\}$, which is not open.
It might be an interesting question to find conditions under which the product of two open subsets is again an open subset. For $\mathbb{R}$ the property is satisfied (Edit: in the comments it is shown for every topological division ring.). And I would be very surprised if it fails for $\mathbb{Z}_p$. More generally, I think that the property holds for many topological rings in practice.