Let $H \leqslant K^{\times}$ be a multiplicative subgroup of the group of units of a field $K$.
It is normal since we're in a field (a commutative ring in particular).
The multiplicative cosets $K/H = \{ xH : x \in K\}$ form a field? Define $xH + yH = (x + y)H$. Then we have the additive identity $0H = \{0\}$ as well as $-xH$ the additive inverse of each $xH$. That is dually to an ideal of a ring where in the quotient we have natural addition and defined multiplication we instead have natural multiplication and defined addition.
Multiplication happens naturally because these are multiplicative subsets: $xHyH = \{ xy h h' : h, h' \in H\} = xyH$ already.
The only problem I see is with $+$ on the cosets being well-defined.
If $a H = bH$, $c H = dH$ then is $(a + c)H = (b + d)H$?
I have so far: If either $a,b,c,d$ are equal to $0$ then we're done. If $x, y \neq 0$, then by definition of multiplicative subset $H$ we have that $xH = yH \iff \dfrac{x}{y} \in H$ where $1/y:= y^{-1}$ etc. If $a = c$ and $b = d$, then there is nothing to prove. Otherwise, say wlog that $b \neq d$. Then $b - d \neq 0$. Then we can muliply the top and bottom of $\dfrac{x}{y}$, where $y = b + d, x = a + c$, by $b - d$ giving the equivalent condition $\dfrac{(a + c)(b - d)}{b^2 - d^2} \in H$ or ...
Suppose $H$ is not the trivial group.
Fix $h\in H$ with $h\ne 1$.
Then letting $$ a=1,\;\;\;b=1,\;\;\;c=-1,\;\;d=-h $$ we get $$aH=bH\;\;\;\text{and}\;\;\;cH=dH$$ but $$(a+c)H=0H\ne (b+d)H$$ so your proposed addition is not well-defined.