Given $a,b\in\Bbb N$ coprime we have that each $ac+bd$ is unique if $0<c<b$ and $0<d<a$ holds.
Given $a,b\in\Bbb N$ coprime it is not true that each $ac+bd$ is unique if $-b<c<b$, $-a<d<a$ and $cd\neq0$ holds.
The number of integers in $$[0,2ab-(a+b)]$$ that get represented by $ac+bd$ when $0<c<b$ and $0<d<a$ holds is $\frac12(a-1)(b-1)-1$ wher $(a-1)(b-1)$ is Frobenius number.
However how many integers in $$[-2ab+(a+b),2ab-(a+b)]$$ get represented by $ac+bd$ if $-b<c<b$, $-a<d<a$ and $cd\neq0$ holds?
Is there a relation to Frobenius numbers?
$ac+bd=ac'+bd'$ happens only if $cc',dd'<0$. Also $c'<0\implies d>0$.
Assume $c'<0$.
$$ac+bd=ac'+bd'\iff a(c-c')=b(d'-d)\iff c=c'+kb\wedge d'=ka+d$$ Moreover $$c=c'+b\wedge d'=a+d$$ holds since $$-b<c,c'<b$$ $$-a<d,d'<a$$ $$cd,c'd'\neq0$$ holds.
There are $2a\cdot2b$ pairs of $c,d$.
Pick $c'<0$. Now for every $d'>0$ we have a $c=c'+b$ and $d=d'-a$ such that $ ac+bd=ac'+bd'$ holds. We have $a$ choices for $d'$ and $b$ choices for $a$.
Consequently $4ab-ab=3ab$ different integers in $[-2ab+(a+b),2ab-(a+b)]$ getting represented.
Is my count correct?