Every problem $P$ can be solved with an oracle machine with oracle access to $P$. In order to get a more meaningful answer, we consider the concept of Turing semi-degree, which is the set of all problems computable with an oracle to $P$, for some problem $P$. The same diagonalization argument used to prove that the halting problem isn't decidable shows that no Turing semi-degree encompasses all languages. (Alternatively, there are uncountably many languages but only countable many problems computable with any given oracle.)
A Turing degree consists of the "hardest" problems in a Turing semi-degree. More formally, it is the set of problems equivalent to some problem $P$: problems which are computable given an oracle to $P$, and conversely $P$ can be computed given an oracle to them. The study of Turing degrees consists a significant part of classical recursion theory.