As user26857 answered the question in a comment, and prefers not to post it as an answer, I'll try to write the answer myself. I think I've understood user26857's argument, but I may be wrong. So, in the lines below, everything that's true is due to user26857, and everything that's false is due to me.
The answer is Yes.
More precisely:
If $A$ is a noetherian integral domain, if $\mathfrak p_1,\dots,\mathfrak p_k$ are distinct nonzero prime ideals of $A$, and if $m$ and $n$ are distinct elements of $\mathbb N^k$, then we have $$\mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}\ne\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}.$$
Proof. In the setting of the question, suppose by contradiction that we have
$$
\mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}=\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}
$$
with $m\ne n$.
Enumerate the $\mathfrak p_i$ in such a way that each $\mathfrak p_i$ is a minimal element of the set $\{\mathfrak p_i,\dots,\mathfrak p_k\}$, and write $\mathfrak p_{ij}$ for the localization of $\mathfrak p_i$ at $\mathfrak p_j$.
For all $i$ we get
$$
(\mathfrak p_{1i})^{m_1}\cdots(\mathfrak p_{ii})^{m_i}=(\mathfrak p_{1i})^{n_1}\cdots(\mathfrak p_{ii})^{n_i}.\quad(1)
$$
Note the following consequence of the determinant trick, or Nakayama's Lemma:
$(2)$ If $\mathfrak a$ and $\mathfrak b$ are ideals of $A$, then the equality $\mathfrak a\mathfrak b=\mathfrak b$ holds only if $\mathfrak a=(1)$ or $\mathfrak b=(0)$.
Let's prove $m_i=n_i$ by induction on $i$:
Case $i=1$: We have $(\mathfrak p_{11})^{m_1}=(\mathfrak p_{11})^{n_1}$ by $(1)$. If we had $m_1\ne n_1$ we could assume $m_1<n_1$, and would get
$$
(\mathfrak p_{11})^{n_1-m_1}(\mathfrak p_{11})^{m_1}=(\mathfrak p_{11})^{m_1},
$$
contradicting $(2)$.
From $i-1$ to $i$: We have
$$
(\mathfrak p_{1i})^{m_1}\cdots(\mathfrak p_{i-1,i})^{m_{i-1}}(\mathfrak p_{ii})^{m_i}=(\mathfrak p_{1i})^{m_1}\cdots(\mathfrak p_{i-1,i})^{m_{i-1}}(\mathfrak p_{ii})^{n_i}.\quad(3)
$$
If we had $m_i\ne n_i$ we could assume $m_i<n_i$ and we could write $(3)$ as
$$
(\mathfrak p_{ii})^{n_i-m_i}\mathfrak b=\mathfrak b
$$
with $(\mathfrak p_{1i})^{n_i-m_i}\ne(1)$ and $\mathfrak b\ne(0)$, contradicting $(2)$. (Here $\mathfrak b$ is the left side of $(3)$, and we assume $2\le i\le k$.) $\square$
Note that the argument still works if $A$ is not noetherian, but the $\mathfrak p_i$ are finitely generated.
Let's call domains having this property $UPIF$-domains, as in your first post.
Observe that a domain which is locally $UPIF$ is $UPIF$.
Indeed, given $$\mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}=\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}$$
we can localize at each of the $\mathfrak{p_i}$ one at a time and use the assumption that $D_{\mathfrak{p_i}}$ is $UPIF$ to deduce that $m_i = n_i$.
From your linked Questions/Answers, we thus have that a locally Noetherian domain is locally UPIF, hence
A locally Noetherian domain is $UPIF$.
However, there are locally Noetherian domains that are not Noetherian.
A classic reference for such examples is section 2 of this paper of Heinzer and Ohm.
For a more recent reference, I'd suggest Loper's article on Almost Dedekind Domains Which Are Not Dedekind. In section 3, he discusses five distinct techniques for producing examples of non-Noetherian domains which are locally discrete valuation rings.
Best Answer
The answer is no.
Let $G$ be the abelian group $\mathbb{Z}\times \mathbb{Z}$ with the lexicographic order, and let $M\subset G$ be the sub-monoid of elements that are greater than or equal to $(0,0)$. Define monoid ideals $I_1:=\{(m,n):m\geq 1\text{ or }n\geq 1\}\subset M$ and $I_2:=\{(m,n):m\geq 1\}\subset M$. For $j\in\mathbb{Z}_{\geq 0}$ and $I\subset M$ an ideal, write $jI:=\{i_1+\ldots+i_j:i_1,\ldots,i_j\in I\}\subset M$. Check that for all $j$, we have $jI_1\neq (j+1)I_1$ and $jI_2\neq (j+1)I_2$, but $I_1+I_2=I_2$.
We get a counterexample to your question by taking $A=k[M]$ to be the monoid ring of $M$ (where $k$ is any field), $\mathfrak{p}_1=k[I_1]$, and $\mathfrak{p}_2=k[I_2]$. Then $\mathfrak{p}_1^j\neq \mathfrak{p}_1^{j+1}$ and $\mathfrak{p}_2^j\neq \mathfrak{p}_2^{j+1}$, but $\mathfrak{p}_1\mathfrak{p}_2=\mathfrak{p}_2$.
An alternate way to describe this example is $A=k[x,y,yx^{-1},yx^{-2},\ldots]$, $\mathfrak{p}_1=(x)$, $\mathfrak{p}_2=(y,yx^{-1},yx^{-2},\ldots)$.