This is not a complete answer, but just an example of relation we can't impose to a $C^*$-algebra.  
  
Let $\mathcal{A}$ be a $C^*$-algebra and let $a, b \in \mathcal{A}$, then the relation $ab-ba = 1$ is impossible :   

**Lemma** : If $ab-ba = 1$ then $ab^n-b^na = nb^{n-1}$.  
*Proof by induction* :   for $n=1$ it's ok.  
Now if it's true for $n$, then $ab^{n+1}-b^{n+1}a = ab^nb -b^{n+1}a = (b^na+ nb^{n-1})b-b^{n+1}a $ $ = b^nab+nb^n-b^{n+1}a = b^n(1+ba)+nb^n-b^{n+1}a = (n+1)b^n $.   $\square$  

**Corollary** : The relation $ab-ba = 1$ is impossible in a Banach algebra.  
*Proof* : If it's possible, then $ab^n-b^na = nb^{n-1}$. Next $\Vert nb^{n-1} \Vert = n \Vert b^{n-1} \Vert  = \Vert ab^n-b^na \Vert $ $ \le \Vert ab^n \Vert + \Vert b^na \Vert \le \Vert ab \Vert \Vert b^{n-1} \Vert +    \Vert b^{n-1} \Vert \Vert ba \Vert$.  
Conclusion $\Vert ab \Vert + \Vert ba \Vert \ge n$  $\forall n$, contradiction.  $\square$     

**Remark** : This relation is realized by unbounded operators.  
For example let the Hilbert space $H = l^2(\mathbb{Z})$, $a: e_n \to ne_{n-1}$ and $b: e_n \to e_{n+1}$