Let $R$ be an integral domain. $r \in R$ has a factorization into irreducibles if there exists a sequence of irreducibles $q_1,\dots,q_n$ such that $r = q_1\dots q_n$. $R$ is a domain with factorization if every non-zero element has a factorization.
Ascending chain condition. Let $X$ be a set and $\leq$ a partial order relation on $X$. $X$ satisifies $a.c.c.$ if for every chain $x_1 \leq x_2 \leq \dots$, the chain stabilizes eventually.
For any integral domain $R$, it is clear that if $R$'s ideals principal ideals satisfying a.c.c. (via inclusion), then $R$ is a domain with factorization. However, intuitively the converse does not hold. If we can find $r = \prod_{i = 1}^{\infty} r_i$ where each $r_i$ is not a unit, and if $r$ also admits a finite factorization, that would be a good example. I cannot find any, though. Could anyone give some hint?
A simple example is the polynomial ring in a countable number of indeterminates over a field $K$, $K[X_1, X_2,\dots, X_n,\dots]$. It is a non-noetherian U.F.D., which is the direct limit of the system of noetherian U.F.D.s $\bigl(K[X_1, X_2,\dots, X_n], i_n\bigr )_{n\ge 1}$, where $i_n$ is the inclusion $K[X_1, X_2,\dots, X_n]\hookrightarrow K[X_1, X_2,\dots, X_{n+1}]$.
Edit:
For the modified question (an integral domain satisfying the a.c.c. for principal ideals which is not a U.F.D.), just take a noetherian domain which is not a U.F.D.
For instance, the ring of integers of a convenient quadratic extension of $\mathbf Q$ which is not principal. Such a ring of integers is a Dedekind domain, and it is principal if and only if it is a U.F.D. Such an example is the ring $\mathbf Z[\sqrt{-5}]$.