Generalisation of chinese remainder theorem on ideals of ring without 1

Question

Let $I_1,\dots,I_n$ be (two-sided) ideals of a ring $R$ (not necessarily with 1), which are pairwise co-maximal, i.e. $\forall i\ne j\in \mathbb{Z}_{[1,n]}$, $I_i+I_j=R$.

Let $f:R\to R/I_1\times R/I_2\times\dots\times R/I_n$, defined by $r\mapsto(r+I_1,r+I_2,\dots,r+I_n)$.

Is $f$ surjective even when $R$ does not have a 1?

Note: In my course, a ring $R$ satisfies:

1. $R$ is an Abelian (commutative) group under $+$.
2. Multiplication is associative.
3. $+$ and $\times$ satisfy distributive laws: $r(s+t)=rs+rt$ and $(r+s)t=rt+st$.

Of course, I would be still interested if you have a proof when $R$ is commutative for $\times$.