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

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).

Conceptually, we can view this as a generalization of the proof that ideals in Euclidean domains are generated by any element of minimal value. The Bezout condition $(1)\Rightarrow$ ideals are closed under gcd, since $\rm\:a,b \in I \:\Rightarrow\: gcd(a,b) = ra+si \in I.\:$ The chain condition $(2)$ says that the divisor relation is well-founded, i.e. that there are no infinite descending chains of proper divisors $\rm\ \cdots\ a_3 \mid a_2 \mid a_1,\:$ which implies ideals can be generated by elements least/minimal w.r.t. divisibility (proof below). Combining $(1)$ and $(2)$ we infer that $\rm\,I\ne 0\,$ is principal, since least generators $\rm\,g_i$ must be associate, else $\rm\:gcd(g_i,g_j) \in I\:$ and it is a proper divisor of $\rm\:g_i,\:$ contra leastness of $\rm\,g_i$ w.r.t. divisibility.

That $\rm\,I\ne 0\,$ can be generated by such minimal generators is provable as follows. First, there exists a set of generators for $\rm\,I,\,$ e.g. the set of nonzero elements of $\rm\, I.\:$ By the chain condition, each generator can be replaced by some divisor $\rm\in I\,$ that is least w.r.t. divisibility, since repeatedly taking proper divisors $\rm\in I\,$ of a generator yields a proper divisor chain $\rm\,\cdots\, a_3 \mid a_2\mid a_1.\:$ This chain must terminate with some $\rm\:a_n\in I\:$ having no proper divisors $\rm\in I\,$ (else it would yield an infinite descending chain of proper divisors, contra the hypothesized chain condition).