I cannot really follow the reasoning you are hinting in your question, but here's my take:
To talk about density you need a topology. Since $M_n(\mathbb{C})$, the space of complex $n\times n$ matrices is finite-dimensional, a very natural notion of convergence is entry-wise; so we can consider the metric
$$
d(A,B)=\max\{ |A_{kj}-B_{kj}|\ : k,j=1,\ldots,n\}, \ \ \ A,B\in M_n(\mathbb{C}).
$$
It is not hard to check that for any matrix $C$,
$$
d(CA,CB)\leq d(A,B)\,\sum_{k,j=1}^n |C_{kj}|,
$$
and the same inequality holds for multiplication on the right (this will be used in the last inequality below).
Now take any $A\in M_n(\mathbb{C})$. Let $J$ be its Jordan canonical form; then there exists a non-singular matrix $S$ such that $J=SAS^{-1}$. Fix $\varepsilon>0$. Let
$$
m=\left(\sum_{k,j=1}^n |S_{kj}|\right)\,\left(\sum_{k,j=1}^n |(S^{-1})_{kj}|\right)
$$
Now, the matrix $J$ is upper triangular, so its eigenvalues (which are those of $A$) are the diagonal entries. Let $J'$ be the matrix obtained from $J$ by perturbing the diagonal entries of $J$ by less than $\varepsilon/m$ in such a way that all the diagonal entries of $J'$ are distinct.
But now $J'$ is diagonalizable, since it has $n$ distinct eigenvalues. And $d(J,J')<\varepsilon/m$. Then $S^{-1}J'S$ is diagonalizable and
$$
d(S^{-1}J'S,A)=d(S^{-1}J'S,S^{-1}JS)\leq m\,d(J',J)<\varepsilon.
$$
We have $A=D_1+N_1=D_2+N_2$ with the given property. And, of course $A$ commutes with $A$, so $A$ commutes with $D_1$ and $N_1$ (and also with $D_2$ and $N_2$). Writing again, $D_1$ commutes with $A=D_2+N_2$, so by the hypothesis again, $D_1$ commutes with both $D_2$ and $N_2$.
Best Answer
Every matrix can be decomposed into Jordan form .
So we have $A=VJV^{-1}$.
Jordan matrix is a sum of some diagonal matrix $D$ and some upper triangular matrix $N$ with $0$'s on diagonal. Because on diagonal there are only zeros all eigenvalues are also $0$.
Hence $N$ is nilpotent (easy proof from CH theorem) - we see that any triangular matrix with 0's along the main diagonal is nilpotent.
We have
$A=V(D+N)V^{-1}$
and because similarity preserves eigenvalues so $N_v=V N V^{-1}$ is also nilpotent.
You can also check that $(V N V^{-1})^k= V N^k V^{-1} $ hence for some $k$ it holds ${N_v}^k=0$.
Commutation.
To $VDV^{-1}$ commute with $VNV^{-1}$ it's enough to show that $D$ commute with $N$ i.e. $ND=DN$. This commutation is assured by the construction of Jordan block.
Take for example $D_i= \begin{bmatrix} \lambda_i & 0 & 0 \\ 0 & \lambda_i& 0 \\ 0 & 0 & \lambda{_i} \end{bmatrix}$ and $N_i= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix}$.
and you will see that $D_iN_i=N_iD_i$.
Multiplication is commutative between these parts of Jordan block because $D_i=\lambda_i I$, consequently multiplication of $N_i$ is just a multiplication of $N_i$ by a scalar $\lambda_i$.