Assuming that the Peano Axioms hold (without the axiom of induction), and assuming one of Robinson's Axioms, namely
Every natural number is either $0$ or the successor of a natural number.
It can be shown that you cannot use the above axiom to prove mathematical induction, since there's an inherent circularity, but I can't seem to pin down what will go wrong.
Let $X$ be the set of polynomials $f(x)$ with integer coefficients which are either $0$ or have positive leading coefficient. This then satisfies your axioms. However, it does not satisfy induction. For instance, if it did, then it would have to satisfy $$\forall a\exists b (a=2b\vee a+1=2b),$$ since you can prove this statement by induction. But this statement is not true for $a=x$.