Let us consider vector space $$V = (\Lambda^2 (\mathbb C(t))^\times)\otimes_\mathbb Z\mathbb Q.$$
Here I consider the group $A=(\mathbb C())^\times$ of non-zero rational functions under multiplication. It is an abelian group and so we can take exterior square.
Denote by $W$ the subspace of $V$ generated by elements of the form $f\wedge g$ such that the divisors of the functions $f$ and $g$ are disjoint (as divisors on $\mathbb P^1$).
Is it true that $V=W$?
Here is an example of some element from $W$:
$$\left(\dfrac{t-a}{t-b}\right)\wedge (t-c)$$
(The points $a,b,c$ are mutually different).
However, it is not clear whether $\xi_{a,b}:=(t-a)\wedge (t-b)$ lies in $W$ or not. (Both divisors of the function $(t-a), (t-b)$ contain $\infty\in\mathbb P^1$.)
On the other hand, it is easy to see that my question is equivalent to the fact that for any $a,b$ we have $\xi_{a,b}\in W$.
Let me write the group $A$ additively and denote $t-a$ by simply $a$. As you mentioned, for $a,b,c$ distinct, $(a-b)\wedge c=a\wedge c-b\wedge c\in W$. In other words, if we work in the quotient $V/W$, then $a\wedge c=b\wedge c$. Using this repeatedly for different permutations of the variables, in $V/W$ we get $$a\wedge c=b\wedge c=b\wedge a=c\wedge a.$$ That is, $a\wedge c=-a\wedge c$ in $V/W$ so $a\wedge c\in W$.
(If you don't tensor with $\mathbb{Q}$, on the other hand, it is not true, since there is a homomorphism $\Lambda^2 A\to\mathbb{Z}/(2)$ which sends $\xi_{a,b}$ to $1$ for all $a\neq b$ and then the kernel of this homomorphism contains $f\wedge g$ for all $f$ and $g$ with disjoint divisors, since if $f\wedge g$ had a sum of an odd number of $\xi_{a,b}$'s then both divisors would have to contain $\infty$.)