Campbell-Baker-Hausdorff formula for three-parameter Lie group I have two operators $A$ and $B$ such that 
$[A,B] = C$
$[A,C] = -2A$
$[B,C] = +2B$
and I would like to obtain an expression for $\log(\exp(A+B)\exp(-B)\exp(-A))$. Is it a linear combination of $A$, $B$ and $C$? If yes, is there a way to calculate explicitly the coefficients?  (I had a look at generalizations of the CBH formula, but this was not very illuminating regarding the second part of my question, i.e. how to calculate the coefficients)
 A: No, Francois, you are wrong that the  $Z\equiv \log(\exp(A+B)\exp(-B)\exp(-A))$ sought is representation-dependent. 
It is, in fact, in the Lie algebra you wrote, here sl(2), so then, indeed, a linear combination of $A$, $B$ and $C$, and representation-independent. 
This is the very essence of the CBH expansion, indeed the foundation of the Lie correspondence ("third theorem"): the most concise, supremely elegant, proof that this logarithm is in the Lie algebra (this linear combination you are asking about) is the one by Eichler, cited in that Crib Sheet. 
Consequently, as detailed, e.g., in R Gilmore's books, and routinely reviewed in remedial Lie theory weeks in physics courses using such expansions, you only need work Z out in a given faithful rep---here, of course, the doublet (Pauli matrices), as you indicated--- and it holds for all faithful representations (injective). 
The point is that the product of group elements is in the group, and the exponential of Lie algebra elements (linear combinations and commutators) will have the exact same commutator coefficients for all reps, since they all obey the same Lie algebraic commutation relations, which you wrote; and, ipso facto, the same combinatorics. (Technically, only the group and the Lie algebra are involved, and not the universal Lie algebra.) 
So, then, you only need find the doublet representation for your algebra, obviously
$$
A=  \begin{pmatrix}
     0&1\\
     0&0
   \end{pmatrix} , \qquad     
B=  \begin{pmatrix}
     0&0\\
     1&0
   \end{pmatrix} , \qquad    
C=  \begin{pmatrix}
     1&0\\
     0&-1
   \end{pmatrix} .     
$$
It is then straightforward to work out the exponentials in the argument of the logarithm, 
$$
\exp(A+B)\exp(-B)\exp(-A)=  \begin{pmatrix}
     \cosh 1 &\sinh 1\\
     \sinh 1&\cosh 1
   \end{pmatrix}  \begin{pmatrix}
     1&0\\
     -1&1
   \end{pmatrix}   \begin{pmatrix}
     1&-1\\
     0&1
   \end{pmatrix}  =  \frac{1}{2}  \begin{pmatrix}
     2/e&e-3/e\\
     -2/e&e+3/e
   \end{pmatrix}   ~   .
$$
This then resolves to the identity plus a Lie algebra element,
$$
=  \frac{1/e+e}{4}1\!\!1 +\frac{1}{4}  \begin{pmatrix}
     -(e+1/e)&2e-6/e\\
     -4/e&e+1/e
   \end{pmatrix} ~.   
$$
The sum is clearly a unimodular matrix; and the second matrix is traceless, and in the Lie algebra; and checks to have traceless elements (think traces of the commutators). 
We normalize it so it is D, s.t. $D^2=1\!\!1$, that is
$$
\frac{1}{4}  \begin{pmatrix}
     -(e+1/e)&2e-6/e\\
     -4/e&e+1/e
   \end{pmatrix}\equiv \frac{\sqrt{e^2+25/e^2-6}}{4}  D. 
$$
Defining 
$$
\frac{\sqrt{e^2+25/e^2-6}}{4}\equiv \sinh \theta ~, 
$$
which checks with the coefficient of the identity above being $\cosh \theta$, we derive 
$$
\exp(A+B)\exp(-B)\exp(-A)= \exp (\theta D).
$$
It  finally follows that 
$$
Z=\frac{\theta}{\sinh \theta }  \begin{pmatrix}
     -(e+1/e)&2e-6/e\\
     -4/e&e+1/e \end{pmatrix}= \\ =  \frac{\theta}{\sinh \theta }   ( -(e+1/e) C  -(4/e)B + (2e-6/e)A  ).
$$
As indicated, the  bottom line expression holds for any faithful representation satisfying your Lie algebra. 
I assume you are interested in physics applications. This is the basic warmup exercise given to students in the first week of review (I hope I didn't do your homework for you). The crib sheet referred to above, in fact, also gives you simple pseudo-differential translation operator examples applying identical CBH combinatorics. 
