How does an oblivious test of plaintext equality work? Assume an ElGamal Cryptosystem. Assume a set of three players, $P_1$, $P_2$ and $P_3$. The private key $x$ is shared among the players. The player $P_1$ has a piece of the private key $x_1$, $P_2$ has the piece $x_2$ and $P_3$ has $x_3$.

Assume the following messages, $m_1$ and $m_2$.

Let $E(m_1)$ and $E(m_2)$ be two ElGamal ciphertexts. 

The elements $r_1$ and $r_2$ are random numbers, and $y = g ^ x$, where $g$ is a generator of the group.

Encrypted message m1: $E(m_1) = (g ^ {r_{1}},   m_1 \cdot y ^ {r_{1}}) = (a_1, b_1)$

Encrypted message m2: $E (m_2) = (g ^ {r_{2}},   m_2 \cdot y ^ {r_{2}}) = (a_2, b_2)$

The question is: how can the holders (players) of the private key execute the oblivious test of plaintext equality proposed by Schnorr and Jakobsson? Specifically, how can the three players can determine if $m_1 = m_2$ without revealing the messages?

PS: The protocol oblivious test of plaintext equality proposed by Schnorr and Jakobsson is in the article entitled "Efficient Oblivious Proofs of Correct Exponentiation".
You've messed up your question. Since the two ciphertexts both use the same r, anyone can easily check if $m_1=m_2$. This is not the interesting case.

But if the two ciphertexts are $(a_1, b_1) = (g^{r_1}, m_1 y^{r_1})$ and $(a_2, b_2) = (g^{r_2}, m_2 y^{r_2})$, then the tuple $(g, y, a_1/a_2, b_1/b_2)$ is of the form the mentioned paper deals with.