$F = (D \times D^{\ast})/\sim = \{[a,b] | b \neq 0\}$ where $[a,b]$ denotes the equivalence class of $(a,b)$. Define addition and multiplication as $[a,b] + [c,d] = [ad + bc, bd]$ and $[a,b][c,d] = [ac, bd]$.
I found the multiplicative identity $[1,1]$. The multiplicative inverse produces the element $[ab, ba]$, is this equivalent to $[1,1]$ because $ab=ba$ and neither one has $b = 0$?
The equivalence relation is $(a,b)$~$(c,d)$ if $ad = bc$.
$$[a,b][b,a]=\cdots\sim[1,1]\qquad\text{because...}$$