If $ab - a'b' = \pm 1$, then can view the linear equation as a linear combination, that leads to the r.h.s. being a multiple of $\gcd$. With that having modulus of $1, \gcd =1$.
Also, the possible co-prime pairs are : $(a,a')=(a,b')=(b,a')=(b,b') =1$. This also means that there two (possibly?) composite number pairs that mutually co-prime, as:
$a = 7, b = 14, a'= 5, b'=15$.
Q.1. Is it possible for $(a,b)=1$, and / or $(a',b')=1$, i.e. for them to be also co-prime; and hence there being 'possibly' $6$ or even $5$ pairs of co-prime numbers.
Coming back, need some integers $x,y$ s.t. $(a+a')x \pm (b+b')y =1$, i.e. the linear combination of $(a+a')$ and $(b+b')$ is $0$ for some integers $x,y$.
I want a solution that is based only on number theory without any recourse to determinants (i.e., effectively linear algebra).
Addendum In wake of answer by @DanielWainfleet, have added link to a simple diagram, that does not help me in improving understanding.
In response to the request (in comments from the proposer) for a "geometric intuition":
Notation : $(x,y)$ is an ordered pair.
If $p=(x,y)$ and $p'=(x',y')$ are points in $\Bbb R^2$ then the area of the triangle with vertices $p,p',$ and $(0,0)$ is $\frac {1}{2}|xy'-x'y|.$ (An area $0$ means that $p,p',$ and $(0,0)$ are co-linear.)
So if $x,y \in \Bbb Z$ and there exists a point $(x',y')\in \Bbb Z^2$ such that the area of the triangle with vertices $(x,y), (x',y'),(0,0)$ is $\frac {1}{2},$ then $|xy'-x'y|=1,$ which implies that $\gcd(x,y)=1.$
If $|ab-a'b'|=1$ then the parallelogram $P$ with vertices $(0,0), (a,b'),(a',b),(a+a',b'+b)$ has area $1$ because it is twice the area of the triangle with vertices $(0,0),(a,b'),(a',b).$ But the triangle with vertices $(0,0),(a+a',b'+b),(a,b')$ also has half the area that $P$ does, so it has area $\frac {1}{2}.$ Applying the preceding paragraph when $a,a',b,b'\in \Bbb Z,$ with $(x,y)=(a+a',b'+b)$ and $(x',y')= (a,b'),$ we have $\gcd (a+a',b'+b)=1.$