I think the answer is no and it follows from the following:

It is consistent that $AC$ fails but for all infinite cardinals $\kappa, 2 \cdot \kappa=\kappa.$

In a model as above, every infinite set is splittable but $AC$ fails in it.

The answer is no and it follows from the following:

It is consistent that $AC$ fails but for all infinite cardinals $\kappa, 2 \cdot \kappa=\kappa.$

The above result is proved by Sageev:

Sageev, Gershon An independence result concerning the axiom of choice. Ann. Math. Logic 8 (1975), 1–184.

In a model as above, every infinite set is splittable but $AC$ fails in it.