Consider a set of operators $$\{A_i\}_{i=1}^N$$ and their commutation relations defined by $$[A_i,A_j]=\Omega_{ij}$$. For the description of my problem, let me introduce the following notation: $$[i,j]\equiv [A_i, A_j]$$ and $$[ij,kl]\equiv [A_iA_j,A_kA_l]$$.

Let's find out what is the commutator $$[ij,kl]$$. Since [i,j] = ij-ji, we can write: $$$$[ij,kl] = ijkl - klij$$$$ I start by transforming $$ijkl$$: \begin{align} ijkl & = [i,j]kl + jikl \\ & = [i,j]kl + j[i,k]l + jkil \\ & = [i,j]kl + j[i,k]l + jk[i,l] + jkli \\ & = [i,j]kl + j[i,k]l + jk[i,l] + [j,k]li + kjli \\ & = [i,j]kl + j[i,k]l + jk[i,l] + [j,k]li + k[j,l]i + klji \\ & = [i,j]kl + j[i,k]l + jk[i,l] + [j,k]li + k[j,l]i + kl[j,i] + klij. \end{align} So, we get $$[ij,kl] = [i,j]kl + j[i,k]l + jk[i,l] + [j,k]li + k[j,l]i + kl[j,i].$$ On the other hand, if we start with $$klij$$: \begin{align} klij & = [k,l]ij + lkij \\ & = [k,l]ij + l[k,i]j + likj \\ & = [k,l]ij + l[k,i]j + li[k,j] + lijk \\ & = [k,l]ij + l[k,i]j + li[k,j] + [l,i]jk + iljk \\ & = [k,l]ij + l[k,i]j + li[k,j] + [l,i]jk + i[l,j]k + ijlk \\ & = [k,l]ij + l[k,i]j + li[k,j] + [l,i]jk + i[l,j]k + ij[l,k] + ijkl. \end{align} So, in this case, we get $$[ij,kl] = -([k,l]ij + l[k,i]j + li[k,j] + [l,i]jk + i[l,j]k + ij[l,k])$$ Now if we set all pairwise commutators to zero, except [i,j], then we get a contradiction, since the first version will give non-zero for the commutator, but the second will give zero. For example, consider the commutator $$[x_0p_0, p_1p_1], ~\textrm{with} ~ [p_i,p_j]=[x_i,x_j]=0 ~ \textrm{and} ~ [x_i,p_j] = i\delta_{ij}$$ Can someone point my mistake as i guess, there should not be any contradiction here.

In your first worked-out example, you first comuuted $$i$$ to the right (past the $$j$$ to the right of it), and then commuted $$j$$ to the right (past the $$i$$ which is now at the rightmost position.
In your third displayed equation, and under the assumption that $$k$$ and $$l$$ commute with everything, you'll find $$[ij,kl] = \underbrace{[i,j]kl}_a + j[i,k]l + jk[i,l] + [j,k]li + k[j,l]i + \underbrace{kl[j,i]}_b\\=\left[[i,j],kl\right]=0\,.$$ Note that only terms $$a$$ and $$b$$ survive, and they cancel each other.
In your procedure, you perform six commutations ($$i$$ past $$j$$, $$k$$, $$l$$ and $$j$$ past $$k$$, $$l$$, $$i$$). But you really only need four of them ($$i$$ and $$j$$ past $$k$$ and $$l$$). In your second computation, you omit the unneccessary $$ij$$ swith and reswitch (but do it for $$kl$$), so you dont't notice the problem.
$$[ij,kl]=[i,j]kl-kl[i,j]=\left[[i,j],kl\right]=\left[j,[kl,i]\right]+\left[i,[j,kl]\right]$$
using the Jacobi identity. So you see that assuming only $$[i,j]$$ to be nonzero also reduces the first result to zero.