Weakness in using only one RSA key pair for two-way communication? In Alice/Bob/Cindy terms (EDIT: and with a little more detail):

Alice and Bob have each securely obtained one key of an RSA keypair from a trusted third party. Alice has one key ($e$ and $n$), Bob has the other ($d$ and $n$, where $d\equiv e^{-1} (mod\ \phi(n))$). 

The RSA algorithm technically does not care which of the two keys is used to encrypt and which to decrypt; if $c_1 \equiv m^e (mod\ n)$ and $m \equiv c_1^d (mod\ n)$, then $c_2 \equiv m^e (mod\ n)$ and $m \equiv c_2^d (mod\ n)$. ($c_1 \neq c_2$)

Therefore, Alice uses $e$ and $n$ to encrypt a message $m_a$ into ciphertext $c_a$ and sends it to Bob, who decrypts it with $d$ and $n$, then encrypts his response $m_b$ with the same $d$ and $n$, and sends the ciphertext $c_b$ to Alice who can decrypt it with $e$ and $n$. Thus, one pair of asymmetric keys is being used to form a two-way communication channel, instead of the normal one-way usage.

Now, Cindy does not know $e$, $d$, or $n$, or any artifact used to produce them, such as $p$ and $q$. she can only see $c_a$ and $c_b$ as they are passed between Alice and Bob. Given that all other security concerns with RSA are properly handled, such as $\geq$2048-bit keys and OEAP padding scheme, is there an "efficient" way (an attack) by which Cindy can obtain $m_a$ and/or $m_b$?
The question and comments seem to be asking the following:  If an implementation of RSA is used in the following way, is it still secure?  An RSA modulus $N = pq$ and exponent $e$ are generated, and (N,e) is given to Party $A$ and $(p,q,e)$ is given to Party $B$.  Then, the parties encrypt their communication where Party $A$ encrypts using the modulus and exponent, while party $B$ encrypts using the private primes and the exponent.

To really answer the question, we'd have to define what encryption padding method is being used (OAEP, PKCSv1.5, etc) and what notion of "secure" we want.  We'd also have to define what it means to encrypt with the secret key instead of the public key.  But as long as the encryption methods with either key are generating the same ciphertext (more precisely, the same distribution over ciphertexts), then whatever reasonable security notion we want will be achieved in this setting under the same hardness assumptions.  This is because the adversary is seeing ciphertexts that are all indistinguishable for ciphertexts generated in the usual way, so the basic security achieved will transfer over easily in provable way.