The definition of an extensive category immediately implies that given two coproduct decompositions indexed by sets of equal cardinality, if the coproduct objects are isomorphic compatibly with their injections, they are isomorphic as coproducts.
Now suppose $\coprod_{i\in I}C_i\cong \coprod_{j\in J}D_j$ are two coproduct decompositions of an object $X$ of an extensive category. By comparing the cardinals $|I|\neq|J|$ we may assume wlog one is strictly greater than the other, and then add to the smaller collections copies of $\mathbf{0}\rightarrow X$ to make the cardinalities of the new pair equal. Then applying the above proposition we see the originally larger collection had redundancies etc. This proves (I think), that the cardinality of nontrivial coproduct decompositions is unique for each object.
Is there any way of proving this without using choice, or comparing cardinals?
No, of course not.
Consider the case with Russell's socks. It is consistent that there is a countable family of pairwise disjoint sets of size $2$, whose union does not have any countably infinite subset.
Namely, the coproduct $\coprod_{n\in\Bbb N}P_n$ is not isomorphic, or even comparable, with $\Bbb N$. But on the other hand, $\Bbb N$ can certainly be divided into countably many sets of size $2$, whose coproduct is indeed $\Bbb N$.
The assumption that cardinal summation extends to arbitrary sums, if so, is not provable without some axiom of choice. While I don't have my books with me here, if my memory serves me right, the question whether or not this implies the axiom of choice is open. So it might as well be that whenever the axiom of choice fails, such examples can be constructed (with index sets perhaps bigger than $\Bbb N$, of course). But it might be strictly between $\sf ZF$ and $\sf ZFC$.