You officially need to first show that $R$ is indeed a ring, before you can go about showing anything involving $R$ is a ring homomorphism. However, if you have a candidate bijection $f$ with another (known) ring, in this case $\Bbb Z$, which means it satisfies $f(x+y)=f(x)\oplus f(y)$ and $f(x\times y)=f(x)\odot f(y)$ as indicated in the answer by rschwieb, then you can cheat a bit because $f$ translates all the operations of $\Bbb Z$ into those of$~R$, and their properties come with them. For instance checking the left distributive law in $R$ can be done as
&=f(x\times y+x\times z)
=f(x\times y)\oplus f(x\times z)\\
&=(f(x)\odot f(y))\oplus(f(x)\odot f(z))
=(a\odot b)\oplus(a\odot c),
where $x,y,z\in\Bbb Z$ are such that $f(x)=a,f(y)=b,f(z)=c$ (which is well defined since $f$ is a bijection). While it looks a bit complicated, it just transfers the responsability for the distributive law from $R$ to $\Bbb Z$ (for which it is used in the middle of the computation). So nothing is really going on. The other axioms can be checked similarly without effort.
This assumes you have a candidate $f$, but also that you know the neutral elements for $\oplus$ and for $\odot$, which must be chosen as $f(0)$ and $f(1)$ respectively (it is part of the requirement for homomorphisms). You can easily deduce from the definitions of $\oplus$ and $\odot$ which these neutral elements are. Once you got them, you know $f(0)$ and $f(1)$, and other values follow, like $f(2)$ which must be $f(1+1)=f(1)\oplus f(1)$. You will see it is downhill from there.