Can you generalise the Chinese Remainder Theorem to noncommutative rings without identity? Ultimately, my question is: does the following theorem hold?

Let $I_1, ..., I_n$ be ideals of some ring $R$, with $R = I_i + I_j$ for $1 \leq i < j \leq n$. 
Then for any $r_1, ..., r_n \in R$ there exists $x \in R$ such that:
$$\begin{align*}
x &\equiv r_1 \pmod{ I_1 }\\ &\vdots \\ x &\equiv r_n \pmod{ I_n}
\end{align*} $$

It's easy enough to show if two such solutions $x$ and $x'$ exist, then $x' \equiv x \pmod{I_1 \cap ... \cap I_n}$. 
Moreover, this shows the ring homomorphism $\phi: R \rightarrow R / I_1 \times \cdots \times R / I_n$, given by $r \mapsto \left ( r + I_1, \ldots, r + I_n \right )$ is surjective. 
Hence, with $\ker(\phi) = I_1 \cap \cdots \cap I_n$, the first isomorphism theorem tells us:
$$ R / \left ( I_1 \cap \cdots \cap I_n \right ) \cong R / I_1 \times \cdots \times R / I_n$$
A few more comments about this question:

*

*Theorem 1.11 of these lecture notes gives a sketch proof with the added condition $R^2 + I_i = R$ for $i = 1, ..., n$.
If $R$ contains a multiplicative identity, then $R^2 = R$, and our requirement immediately holds. Is this added condition necessary for $n > 2$?

*Judson's Abstract Algebra textbook poses the $n=2$ case as an exercise with no added assumption about multiplicative identities etc... which is answered in this question.

 A: Ultimately, my question is: does the following theorem hold?
No. For a counterexample, start with the unital ring $\mathbb Z_2\times \mathbb Z_2$ and change the usual multiplication to one where $xy=0$ for every $x$ and $y$. That is, let $R$ be the ring that has additive structure equal to that of the $4$-element noncyclic group and multiplication that is constant zero.
This ring $R$ has three $2$-element ideals, $I_1=\mathbb Z_2\times \{0\}$, $I_2=\{0\}\times \mathbb Z_2$, $I_3=\{(x,x)\;|\;x\in \mathbb Z_2\}$. These ideals are comaximal (i.e., they satisfy $I_m+I_n=R$ when $m\neq n$). But there is no surjective homomorphism from $R$ to $R/I_1\times R/I_2\times R/I_3$, since $R$ has $4$ elements and $R/I_1\times R/I_2\times R/I_3$ has $8$ elements. For a specific failure of the CRT, choose any $r\in R-I_3$. There exists no solution to

$$x \equiv 0 \mod I_1 \\ x\equiv 0\mod I_2 \\ x \equiv r \mod I_3$$

because $I_1\cap I_2\subseteq I_3$.
The additional (necessary and sufficient) hypothesis that you need on the ideals $I_1,\ldots,I_n$, beyond comaximality, to derive the CRT for nonunital rings is that the ideals $I_1,\ldots,I_n$ generate a distributive sublattice of the ideal lattice of $R$. Checking this for a finite set of pairwise comaximal ideals  is equivalent to checking that for each $j$, the ideal $I_j$ is comaximal with $\bigcap_{k\neq j} I_k$. You can see how this condition fails in the example above: e.g. $I_3$ is not comaximal with $I_1\cap I_2 = (0)$.
