After some thought, I think the answer is in fact NO, even for CCA1 and even for Shoup's OAEP+.

RSA-OAEP/OAEP+ work by taking a message $m$, producing a padding $p(m,r)$ and then encrypting this, so $c = f(p(m,r))$ where $f$ is RSA encryption, and $f(u) = u^e \pmod{N}$ is deterministic. In fact, the whole point of OAEP(+) is to inject some entropy into ciphertexts which is required for IND-CPA and higher security.

ElGamal encryption is already randomised. If we try ElGamal-OAEP(+) we get $c = (g^r, y^r \cdot p(m, r'))$ where $y$ is the public key. Since ElGamal is homomorphic, this is obviously not even CCA1: consider an adversary who picks $m_0, m_1$, asks for a challenge ciphertext $c = (u, v)$ and then sets $c' = (u \cdot g^s, v \cdot y^s)$ for randomly chosen $s$. This is still a valid OAEP(+) ciphertext whatever the padding $p$ is (since we're only changing the "outer" randomness $r \mapsto r + s$) so the CCA1 game will happily decrypt this and return $m_0$ or $m_1$ as desired.

This is assuming of course that you can map your padding function's range into the group over which you're doing ElGamal --- for ECC, this should be fine, for $\mathbb Z^\times_p$ groups it's harder. As an alternative one could consider hashed ElGamal-OAEP+ with $c = (g^r, H(y^r) \oplus p(m, r'))$ where $H$ is independent of the hash functions used in the OAEP+ padding $p$. My intuition is that this is still not CCA1, even though it doesn't have the homomorphic property anymore. Certainly if $H$ has some homomorphic properties itself then one should be able to do something like the above counterexample.

After some thought, I think the answer is in fact NO, even for IND-1-CCA^* and even for Shoup's OAEP+.

RSA-OAEP/OAEP+ work by taking a message $m$, producing a padding $p(m,r)$ and then encrypting this, so $c = f(p(m,r))$ where $f$ is RSA encryption, and $f(u) = u^e \pmod{N}$ is deterministic. In fact, the whole point of OAEP(+) is to inject some entropy into ciphertexts which is required for IND-CPA and higher security.

ElGamal encryption is already randomised. If we try ElGamal-OAEP(+) we get $c = (g^r, y^r \cdot p(m, r'))$ where $y$ is the public key. Since ElGamal is homomorphic, this is obviously not even CCA1: consider an adversary who picks $m_0, m_1$, asks for a challenge ciphertext $c = (u, v)$ and then sets $c' = (u \cdot g^s, v \cdot y^s)$ for randomly chosen $s$. This is still a valid OAEP(+) ciphertext whatever the padding $p$ is (since we're only changing the "outer" randomness $r \mapsto r + s$) so the IND-1-CCA game will happily decrypt this and return $m_0$ or $m_1$ as desired.

This is assuming of course that you can map your padding function's range into the group over which you're doing ElGamal --- for ECC, this should be fine, for $\mathbb Z^\times_p$ groups it's harder. As an alternative one could consider hashed ElGamal-OAEP+ with $c = (g^r, H(y^r) \oplus p(m, r'))$ where $H$ is independent of the hash functions used in the OAEP+ padding $p$. My intuition is that this is still not CCA1, even though it doesn't have the homomorphic property anymore. Certainly if $H$ has some homomorphic properties itself then one should be able to do something like the above counterexample.

_{IND-1-CCA: Is standard IND-CCA2 where you only get 1 decryption query after seeing the challenge ciphertext instead of polynomially many.}