I think some proofs for unital rings carry through without change, but maybe I'm being stupid? For example:
As Martin noted in comments, we can assume $n=2$, so consider an ascending chain
$$I_1\leq I_2\leq \dots$$
of ideals of $A\times B$, where $A$ and $B$ are Noetherian. I'll identify $A$ with the ideal $A\times\{0\}$ of $A\times B$, and let $\pi:A\times B\to B$ be the projection map.
Then
$$I_1\cap A\leq I_2\cap A\leq\dots$$
is an ascending chain of ideals of $A$, and
$$\pi(I_1)\leq\pi(I_2)\leq\dots$$
is an ascending chain of ideals of $B$.
Since $A$ and $B$ are Noetherian, there is some $t$ such that $I_i\cap A=I_t\cap A$ and $\pi(I_i)=\pi(I_t)$ for all $i\geq t$.
But if $(a,b)\in I_i$ for $i\geq t$, then $b\in\pi(I_i)=\pi(I_t)$, so
$(a',b)\in I_t$
for some $a'\in A$. So
$$(a-a',0)=(a,b)-(a',b)\in I_i\cap A=I_t\cap A,$$
and so
$$(a,b)=(a-a',0)+(a',b)\in I_t.$$
So $I_i=I_t$.
There are examples in the rings you're looking at. For instance, consider the ideal generated by $(4,4)$ in $2\mathbb{Z}\oplus2\mathbb{Z}$. This consists of all $(x,y)$ such that $x$ and $y$ are both divisible by $4$ and $x-y$ is divisible by $8$. In particular, it contains $(4,4)$ but not $(4,0)$, so it cannot have the form $I_1\oplus I_2$.
For an example that is a bit simpler to check, let $A$ be any nontrivial abelian group and make it a ring by defining $ab=0$ for all $a,b\in A$. Then $\{(a,a):a\in A\}$ is an ideal in $A\oplus A$ that does not have the form $I_1\oplus I_2$.
Best Answer
define$$\phi : R_1\times R_2 \to R_1/I_1 \times R_2/I_2$$ such that $$\phi (a,b)= (a+I_1 , b+I_2)$$ then $$ker\phi = I_1\times I_2 .$$ Now use First isomorphism theorem