It is important to keep in mind where the different elements are. If $f:\Bbb Z \to R$, then there is no sense in applying $\oplus$ and $\odot$ to the arguments of $f$. Writing something like $f(a\odot b)$ is not very productive, since $a, b$ are considered to be elements of $\Bbb Z$, and not elements of $R$ where these operations makes sense. Rather, you want to apply those operations to the resulting elements, like so: $f(a)\odot f(b)$.
The fact that they are both rings defined on the set of integers (showing that $R$ is in fact a ring is important as well) does make it difficult to keep $\Bbb Z$ and $R$ apart, but it is necessary to do so in order to understand what is going on. If you want to show to show that $f$ is a homomorphism, you want to show the following:
- $f(a + b) = f(a)\oplus f(b)$
- $f(ab) = f(a)\odot f(b)$
Remember, if $a, b\in \Bbb Z$, then $\oplus$ and $\odot$ have no business being near them. Similarily, if $c, d \in R$, then you need to be very careful about $+$ and $\cdot$, and remember that they are only a means for calculating $\oplus$ and $\odot$, not "real" operations.
I will do point 1 above, I hope you can follow it to show number 2.
f(a + b) = -(a + b) + 1 = -a -b + 1 = (-a + 1) + (-b + 1) - 1 \\= f(a) + f(b) - 1= f(a) \oplus f(b)
Here is a small addendum on how one might deduce $f$. First of all, we need to identify the additive and multiplicative identities $0_R$ and $1_R$. That can be done by solving the equations $a\oplus 0_R = a$ and $b \odot 1_R = b$. This will result in $1_R = 0$ and $0_R = 1$.
Now, assume there is a (non-trivial) homomorphism $f$. We will use $f(1) = 0$ to deduce what $f(a)$ must be for a general $a$ if it does exist. We can start by deducing $f(2)$:
f(2) = f(1+1) = f(1)\oplus f(1) = 0\oplus 0 = 0 + 0 - 1 = -1
Now $f(3)$ and $f(4)$ are calculated similarily to be $-2$ and $-3$, and a pattern seems to emerge; the definition $f(a) = -a + 1$ fits so far. To show that this actually is a homomorphism, we prove the two steps above, and since any homomorphism from $\Bbb Z$ is completely determined by what its value on $1$ is, this is the only one.
Once we show that it is bijective, we know that it is an isomorphism, and therefore, $\Bbb Z$ and $R$ are isomorphic.