No, this property does not imply that $A$ is locally Noetherian.
For example, let $F\subset L$ be an extension of fields, and let $A=F+XL[[X]]$ (that is, $A$ is the set of power series over $L$ whose constant term belongs to $F$). Then, $A$ is a one-dimensional local domain with maximal ideal $\mathfrak{m}=XL[[X]]$. Hence, any product of nonzero prime ideals is equal to $\mathfrak{m}^n$ for some integer $n$, and these are all distinct since $\mathfrak{m}^n=X^nL[[X]]$, so your property holds.
However, $A$ is not necessarily Noetherian: indeed, if $\mathfrak{m}$ is finitely generated (as an ideal over $A$) then $L$ will have a finite generating set as an $F$-vector space. In particular, if $[L:F]=\infty$ then $A$ is not Noetherian (and so, being local, it is neither locally Noetherian).