Is there a way to construct a quantum circuit/oracle to check if 2 qubits in an unknown pure state are entangled?

No.

For instance, if I either give you $|00rangle$ or $|11rangle$ with 50% probability each, or $|00ranglepm|11rangle$ with 50% probability each, there is no way to distinguish these two cases - not even with any whatsoever small probability. The mathematical reason is that those are described by the same density matrix - but you always get some pure state!

However, one of those sets consists of entangled states, while the other one doesn't.

The situation is different if you get the same state several times and can do a number of tests on each of those copies.

$newcommand{calU}{mathcal{U}}newcommand{ket}[1]{lvert#1rangle}$Suppose such a circuit existed, and denote with $calU$ the unitary describing its overall action. For this to be a "circuit detecting entanglement", there should be two types of output states, one corresponding to the answer "yes, the input was entangled" and the other corresponding to the answer "no, the input was not entangled".

In other words, there should be two orthogonal subspaces $mathcal V_e$ and $mathcal V_{ne}$ such that $calUlvertpsirangleinmathcal V_e$ for all entangled states $lvertpsirangle$, and $calUlvertphirangleinmathcal V_{ne}$ for all separable $lvertphirangle$.

Let $ket{psi_1},ket{psi_2}$ be two orthogonal separable states, so that $calUket{psi_i}inmathcal V_{ne}$. Then, by linearity, we must also have $calU(alphaket{psi_1}+betaket{psi_2})inmathcal V_{ne}$ for any pair of complex coefficients with $|alpha|^2+|beta|^2=1$.

But all entangled states can be written as superpositions of separable states, thus this circuit is bound to incorrectly classify all entangled states as separable as well.