Is there a way to remove elements from $\Bbb{Z}$ and create a related ring structure? For example.  Everyone hates $3$, so let's remove it all-together from $\Bbb{Z}$: Let $\Bbb{Z}' = \Bbb{Z} - 3\Bbb{Z}$.    Then is there a way to keep it a ring, i.e. $\Bbb{Z}'$ forms a ring by redefining $+$ somehow?  By that I mean not too much as obviously any infinite set can be turned into an additive group.  But for instance, something like, if $a, b \in \Bbb{Z}', a + b \in 3\Bbb{Z} \implies a + b := 0$.  Something simple like that.  
 A: Here is a (rather clunky) way of putting an additive structure on ${\Bbb Z}-3{\Bbb Z}$:
$$m\oplus n=\Bigl\lceil\frac{m+n+1}{3}\Bigr\rceil
  +\Bigl\lceil\frac{2m+2n+2}{3}\Bigr\rceil-1\ .$$
Doesn't make a ring structure with ordinary multiplication though, you need to redefine multiplication too, and I think it will look even worse than addition ;-)
A: Let $p_i, i=1...n$ be a subset of the set $P$ of all primes such that that the set of the other primes $P'=P-\{p_i, i=1...n\}$ is infinite. Let $\mathbb Z'=\mathbb Z-\cup \mathbb p_iZ$. Then you can turn $\mathbb Z'$ into a ring while retaining $\mathbb Z$'s multiplication structure and redefining $+$ as follows:
Since $P$ and $P'$ have the same cardinality there exists a bijection $f: P \to P'$. Extend $f$ to map $0\to 0$, $1\to 1$, and to all integers by mapping $\Pi_j p_j^{k_j}\to \Pi_j f(p_j)^{k_j}$. This obviously preserves multiplication structure.
Then introduce the addition operation: for $a,b\in \mathbb Z'$ define $a+_{\mathbb Z'}b=f(f^{-1}a+_{\mathbb Z}f^{-1}b)$.
It only remains to check whether $\times_{\mathbb Z'}$ is associative relative to $+_{\mathbb Z'}$:
$(a+_{\mathbb Z'}b)\times_{\mathbb Z'} c=f(f^{-1}a+_{\mathbb Z}f^{-1}b)\times_{\mathbb Z'} f(f^{-1}c)=f((f^{-1}a+_{\mathbb Z}f^{-1}b)\times_{\mathbb Z}f^{-1}c)=f(f^{-1}a\times_{\mathbb Z}f^{-1}c+_{\mathbb Z}f^{-1}b\times_{\mathbb Z}f^{-1}c)=f(f^{-1}(a\times_{\mathbb Z'}c)+_{\mathbb Z}f^{-1}(b\times_{\mathbb Z'}c))=a\times_{\mathbb Z'}c+_{\mathbb Z'}b\times_{\mathbb Z'}c$
