Public key homomorphic encryption based on secret key homomorphic
What you found out is really close to the result you want. In fact, your $SOMETHING$ will be $m_1R_2+m_2R_1+R_1R_2$. 

The only missing step is giving out an encryption of $SOMETHING$ as the first component of your ciphertext.

In order to do so, notice that you can easily multiply $a_1$ by $b_2$, since $b_2$ is a bit: 
$a_1 \times b_2 = a_1$ if $b_2 = 1$, and $0$ otherwise. (this $0$ can be computed as $a_1⊕a_1$, and $b_2$ is a public value) 

Now, $a_1 \times b_2$ encrypts $m_2 R_1 + R_1 R_2$. By symmetry, you can remove $a_2 \times b_1$ that encrypts $m_1 R_2 + R_1 R_2$. 

Finally, notice that $a_1 \times a_2$ encrypts $R_1R_2$.

We are now ready to give $a_1\times b_2 + a_2\times b_1 - a_1 \times a_2$ which encrypts $m_2R_1+R_1R_2+m_1R_2+R_1R_2-R_1R_2=SOMETHING$.

Your final homomorphic multiplication for your public key scheme is thus:

$$(a_1,b_1)\times(a_2,b_2) = (a_1\times b_2 + a_2\times b_1 - a_1 \times a_2,\, b_1 \times b_2)$$