Does this ring of proper fractions have another description? Let S be the ring of rational numbers in $[0,1)$ with operations defined by $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc \:(mod \: bd)}{bd}$ and $\frac{a}{b}\times\frac{c}{d}=\frac{ac \:(mod \: bd)}{bd}$.  Basically its a way to make a ring out of proper fractions.  I've never seen this ring before, and I was lead to it by considering the quotient ring $R/(x)$ where $R=x\mathbb{Q}[x]+\mathbb{Z}\subset\mathbb{Q}[x]$.  Is S isomoprhic to anything more familiar?   
 A: It's not a ring.  For instance, $$\frac{1}{2}\cdot\left(\frac{1}{2}+\frac{1}{2}\right)=\frac{1}{2}\cdot0=0$$ but $$\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}.$$
A: As Eric Wofsey has already shown, this does not describe a ring.
To go back to your motivating example of $R/(x)$ with $R=\mathbb Z+x\mathbb Q[x]$, we can also clarify what that should look like.
Your description in the comments is apt: two things are equal if they have the same constant term, and if their 'linear' term is equal modulo $\mathbb Z$.  That suggests a ring structure on $\mathbb Z\oplus \mathbb Q/\mathbb Z$.
Now, there is a well known ring structure on $\mathbb Z\oplus \mathbb Q/\mathbb Z$: it is called the idealization of $\mathbb Q/\mathbb Z$ by $\mathbb Z$. The sum is given by 
$(n,q\mod{\mathbb {Z}})+(m,p\mod{\mathbb {Z}})=(n+m, q+p\mod{\mathbb {Z}})$ and the product is $(n,q\mod{\mathbb {Z}})(m,p\mod{\mathbb {Z}})=(nm,np+mq\mod{\mathbb {Z}})$.
You should be able to verify that the homomorphism $\sum a_ix^i\mapsto (a_0, a_1\mod \mathbb Z)$ from $R\to \mathbb Z\oplus \mathbb Q/\mathbb Z$ has kernel $(x)$.
I can't be sure what led you to the candidate ring in your question, but it "sounds" like you were considering the subset with constant terms zero. In this idealization, $\mathbb Q/\mathbb Z$ does become a ring, but the multiplication is uniformly zero.
