The answers to the two parts of the first part of the question are "no" and "probably yes". Take the free associative algebra (over any ring, say, $\mathbb{Z}$) with free generators $x,y$. Then $x,y$ do not commute in your sense. The main obstacle is that $q_1,q_2$ in your definition must really commute and commuting elements in free rings have been described by Bergman. The description is very similar to the description of commuting elements in a free group/semigroup, it immediately implies that $q_1,q_2$ satisfying your conditions do not exist.

For every $c_1,c_2$, there probably exists a bigger ring, where $c_1,c_2$ "relaxly commute". Denote the initial ring by $C$. Your question is equivalent to the following one. Consider the ring $R$ given by the generators of $C$ plus $q_1,q_2$ and relations (1),(2),(3) plus all relations of $C$, and ask whether the natural homomorphism $\phi$ from $\langle c_1,c_2\rangle$ into $R$ is injective. I will prove it for the case when $C=\langle c_1, c_2\rangle$ is free, i.e. has no non-trivial relations (the general case should be similar, but more difficult). Notice that the relations (1), (2), (3) do not have any overlaps (use the order $c_1 < c_2 < q_1 < q_2$). Thus these relations form a (non-commutative) Groebner-Shirshov basis of $R$. Since every leading term in your relations contains $q_2$, no non-zero (non-commutative) polynomial in $c_1, c_2$ is equal to 0 in $R$, hence $\phi$ is injective. For general $C$, I would recommend reading this paper.

About the second part of the question. The answer is "no" (which also gives the "no" answer to the first part of the first part of the question). The matrices $\left(\begin{array}{ll} 1 & 2 \\\ 0 & 1\end{array}\right), \left(\begin{array}{ll} 1 & 0 \\\ 2 & 1\end{array}\right)$ do not relaxly commute in $M_2(\mathbb{C})$. In fact conditions (1) and (3) cannot be satisfied together for these matrices. It is easy to see: these conditions can be rewritten as a system of linear equations with entries of $q_1,q_2$ as unknowns, and this system is inconsistent.

** Update 1. ** Here is the Maple program that proves inconsistency:

restart;with(linalg):

A:=matrix([[1,2],[0,1]]); B:=matrix([[1,0],[2,1]]):

C:=evalm(A&*B-B&*A):

P:=matrix([[a,b],[c,d]]): Q:=matrix([[x,y],[z,t]]):

W:=evalm(A&*P-B&*Q-C):

V:=evalm(A&*Q-Q&*A-B&*P+P&*B):

zz:=solve({W[1,1]=0, W[1,2]=0, W[2,1]=0,W[2,2]=0,V[1,1]=0,V[1,2]=0}):

subs(zz,evalm(V));

Output:

$$ \left[ \begin {array}{cc} 0&0\\\ 16&0\end {array}
\right]$$

instead of 0.

** Update 2. ** The matrices $\left(\begin{array}{lll} 1 & 1 & 0\\\ 0 & 1 & 0\\\ 0 & 0 & 1\end{array}\right), \left(\begin{array}{lll} 1 & 0 & 0\\\ 0 & 1 & 1\\\ 0 & 0 & 1\end{array}\right)$
relaxly commute. One can take $q_1=\left(\begin{array}{lll} 0 & 0 & -1\\\ 0 & 0 & 0\\\ 0 & 0 & 0\end{array}\right), q_2=\left(\begin{array}{lll} 0 & 0 & 0\\\ 0 & 0 & 0\\\ 0 & 0 & 0\end{array}\right)$.

The Lie algebra generated by these matrices (inside the Lie algebra of all $3\times 3$-matrices) is of dimension 4 spanned by the identity matrix and three matrix units $E_{i,j}$, $1\le i < j\le 3$. So it is the direct product of the Heisenberg Lie algebra and the 1-dimensional Lie algebra.