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.