is $E(x),E(y) ↦ E(x+y)$ well defined? 
Let $x,y\in \mathbb R$ and $x\sim y \iff x-y\in \mathbb Z$. $E(x)$ is
  the equivalance class containing $x$.

a) Is $E((x),E(y)) ↦ E({x+y})$ well defined?
Where $\rightarrow$ means an operation
b) Is $(E(x),E(y)) ↦ E(xy)$?

Yes it is well defined.  Here is the proof.
Fix some $x,y\in\mathbb{R}$, and choose some $x'\in E(x)$ and $y'\in E(y)$.  It suffices to show that $E(x+y)=E(x'+y')$.
This is easy to see, as
$$(x'+y')-(x+y)=(x'-x)+(y'-y)=\text{integer}+\text{integer}=\text{integer}.$$
Now consider the case of multiplication.  We would want to show that $E(xy)=E(x'y')$.  However this is not always the case.  Observe that
$$x'y'-xy=(x'y'-x'y)+(x'y-xy)=x'(y'-y)+y(x'-x)=x'\cdot\text{integer}+y\cdot\text{integer}.$$
We need only choose any example in which the latter formula is not an integer.  Take for example
$x=0$
$x'=1$
$y=\pi$
$y'=\pi+1$.
