Prove Principal Ideal Domain from Bezout's condition, and terminating divisibility chain

The second condition is to assure that every ideal has a finite set of generators. It is an ascending chain condition for principal ideals. Let $I\subseteq R$ be any ideal and start with $a_0:= 0$. In each step, pick $a'_i\in I\setminus \langle a_{i-1}\rangle$ (an element which can not be generated by $a_i$) and set $a_i:=\gcd(a_{i-1},a'_i)$. By the first assumption, $a_i\in I$ and clearly $\langle a_{i-1}\rangle \subseteq \langle a_i \rangle$. Since the $a_i$ form a chain as in (2), we may now conclude that this process actually ends at some point $a_N$, at which we have found the generator $a:=a_N$.

Other answers have already shown why you need the second condition. As a counterexample, consider the ring of all algebraic integers, that is the set of complex numbers which satisfy a monic polynomial over $\mathbb{Z}$. This ring satisfies your first condition, but isn't even a unique factorisation domain, let alone a principal ideal domain (this ring has no irreducible elements, since the square root of an algebraic integer is itself an algebraic integer).