No, but the counterexample is trivial. Take any integral domain with finitely many prime ideals which is not a field. For example, the localization $\mathbb{Z}_{(p)}$ of the integers at a prime p. The zero ideal is non-maximal and prime so, trivially, $\prod_{i=1}^np_i=0$. Maybe this isn't exactly what you were meaning to ask?
Induct on $r$. If $r=1$, then done.
Let $r=2$, then $A'\subset P_1\cup P_2.$ We can assume $A'\nsubseteq P_1$ and $A'\nsubseteq P_2$. Let $x_1\in A'$ and $x_1\notin P_1$. Similarly $x_2\in A'$ and $x_2\notin P_2$. If $x_1\notin P_2$, we get a contradiction that $A'\subset P_1\cup P_2$. So we have $x_1\in P_2$. Similarly $x_2\in P_1.$ Then $x_1+x_2\in A'$, but $x_1+x_2\notin P_1\cup P_2$.
Next let $r\geq 3$. Now assume the statement for $r-1$. That is for any choice of $r-1$ ideals out of which atmost two of them are not prime ideals, then the result holds.
If $A'$ is contained in any of the $r-1$ many ideals out of $P_1,\dots, P_r$, then we are done by induction. So assume otherwise. We can safely assume that $P_r$ is a prime ideal.
Let $x\in A'$ such that $x\notin P_j$ for all $j=1,\cdots r-1$. Now if $x\notin P_r$ , then $x\notin \cup_{j=1}^r P_j$, a contradiction. So we can assume that $x\in P_r$.
Note $A'P_1\cdots P_{r-1}\nsubseteq P_r$, for otherwise either of the factors is contained in $P_r$, in that case we are done by induction. Let $y\in A'P_1\cdots P_{r-1}$ and $\notin P_r$. Now $x+y\in A'$, such that $x+y\notin P_i$ for all $i.$
So we are done.
Best Answer
Using the result mentioned by YACP in the comments, the problem reduces to proving: If an ideal $I$ contained in a union of prime ideals $P_i$, then $I$ is contained in one of the $P_i$.
We can prove that result by induction on $n$. The case $n=1$ is clear. For the inductive step, if $I$ is contained in a union of less than $n$ of the $P_i$, then the result holds by induction. Otherwise, for every $i$ there exists an element $x_i \in I$ which is in $P_i$ but not in any $P_j$ for $j \ne i$. Then the element $x = x_1 + x_2 x_3 \ldots x_n$ is in $I$, but cannot be in any of the $P$'s (why not?): contradiction.
(Do you see where we used the fact that the $P_i$ are prime?)