Non commutative topological manifolds 
Assume  that $A$ is  a  Banach  algebra with two closed two sided ideals  $I$  and  $J$ such that $I$  and  $J$ are  commutative  and $A=I+J$. Does this implies that $A$ is  commutative? For the  $C^{*}$  algebra, the  answer is  "Yes".

The motivation: A (noncommutative) compact n dimensional topological  manifold could  be  defined as  follows:
A  (non  commutative)  $C^{*}$ or Banach algebra  $A$  such that there  are ideals  $I_{k}$, $k=1,2,\ldots,n$ such that $A=I_{1}+I_{2}+\ldots I_{k}$ and  each $I_{j}$ is  isomorphic to $C_{0}(\mathbb{R}^{n})$. 
But  the  above  answer in MSE shows that, in the context of  $C^{*}$  algebras, this  definition does  not  give  any non  commutative  example,.

So  we  search  for  a non commutative  example  in  Banach  algebras.

Note: According to the comment on my MSE question: To what extent  Banach or  $C^{*}$  algebras whose underline Lie  algebras are  metabelian  are studied and  classified?
 A: Theorem Let $A$ be a unital ring and $I_1,\dots,I_n \subset A$ be 2-sided commutative ideals such that $A=I_1+\dots + I_n$. Then, $A$ is commutative.
Proof: If $A=I_1+\dots+I_n$, then $1 = x_1+\dots+x_n$ for $x_i \in I_i$. But then,
$$1 = (x_1+\dots+x_n)^{n+1} \in I_1^2 +\dots+ I_n^2$$ and we conclude that $A=I_1^2 + \dots + I_n^2$.
Now, if $I \subset A$ is any abelian 2-sided ideal, then $I^2$ is central, since
$$a(bc) = (ab)c= c(ab) = (ca)b = b(ca) = (bc)a,$$
for any $a \in A$ and $b,c \in I$.
Since $A = I_1^2 + \dots + I_n^2$, $A$ is commutative. q.e.d.
If $A$ is not unital, then the same argument still shows that $A^{n+1} \subset A$ is central. If $A^2 \neq A$, then there are counterexamples - even for Banach algebras. For example, one can consider the non-abelian algebra $$A = \left( \begin{matrix} 0 & * & * \\ 0 & 0 & *\\0 & 0 & 0 \end{matrix}\right)$$ of strictly upper-triangular $3\times 3$-matrices (with entries in a field), which is is a sum of two abelian ideals
$$I = \left( \begin{matrix} 0 & 0 & * \\ 0 & 0 & *\\0 & 0 & 0 \end{matrix}\right), \quad J = \left( \begin{matrix} 0 & * & * \\ 0 & 0 & 0\\0 & 0 & 0 \end{matrix}\right).$$
