Fortunately, the obvious idea works out well in this case.
Letting $S$ denote $\oplus_{i\in \Bbb Z}R_i$, we might speculate that everything in the subring generated by $S$ and $1_R$ looks like $T=\{x+n1_R\mid x\in S, n\in\Bbb Z\}$.
I don't think you'll have any trouble verifying that it's a ring: it's clearly closed under addition and multiplication, and has the same additive and multiplicative identities as the product ring above it. (Be sure to carry this verification out!)
Certainly the ring $T$ contains both $S$ and $1_R$, and conversely it's easy to see that any ring containing $S$ and $1_R$ must contain all the elements of $T$. So, $T$ is the ring generated by $S$ and $1_R$.
One last task: I mentioned that the obvious idea works this time. Suppose $R_1$ and $R_2$ are subrings of $R$, and try a similar idea to the one we used above with these rings. Can you see why the old idea doesn't directly work, and that something more complex is needed?