First of all apologies for my "hands on" language with coordinates and indices, I am aware that this is not how Lie algebras are normally discussed, but it is the language I'm familiar with so please bear with me.
Say I have a real Lie algebra g, of which I know the brackets in a basis $t^i$: $$ [t^i , t^j ] = c^{ij}{}_k \,t^k $$ and I know the multiplication, inverse and identity maps of the (conncected component of the identity of the) corresponding Lie group G, through the exponential map with a certain ordering, for example ($x_i,y_i,m_i\in \mathbb{R}$): $$ e^{x_i t^i} e^{y_i t^i} = e^{m_i(x,y) t^i} $$ (in this case I chose the symmetric ordering but it's not important, I could have factorized it any other way), and similarly for the other maps.
Now, say I introduce a central extension of the algebra, call it g': $$ [t^i , t^j ] = c^{ij}{}_k \,t^k + s^{ij} z \,, $$ where $z$ is a new central generator, $[z,t^i]=0$, and $s^{ij}$ is an antisymmetric matrix that satisfies the 2-cocycle condition: $$ c^{ij}{}_l s^{kl} + c^{jk}{}_l s^{il} + c^{ki}{}_l s^{jl} = 0 \,, $$ so that the Jacobi identities of the new algebra are satisfied.
My question is, is there a closed formula to express the multiplication map of the centrally-extended Lie group G' in terms of the product map of G?
I know, from the BCH formula, that the product of two exponentials of elements of g' will have the form: $$ e^{a \,z+x_i \,t^i} e^{b\,z+y_i\, t^i} = e^{ [a+b + s^{ij} h_{ij}(x,y)] z + m_i(x,y) \, t^i} $$ where $h_{ij}(x,y)$ is a map that must satisfy a compatibility condition with $m_i$(coming from associativity), i.e. $$ h_{ij}(x,m(y,u)) = h_{ij}(m(x,y),u) \,, $$ moreover, $h_{ij}(x,y)$ must be zero when either $x_i$ or $y_i$ are zero, and must be zero when $x_i = - y_i$ (or whatever the inverse map is, in case we didn't use the symmetric ordering).
I feel like one should be able to write the map $h_{ij}(x,y)$ explicitly in terms of the map $m_i$, but so far I haven't been able to find out how. is my impression wrong? Is there anything more I can say about the map $h_{ij}$ besides the general properties I just listed?
Edit: of course I can calculate $h_{ij}(x,y)$ as a series in powers of $c^{ij}{}_k$ using the BCH formula, I get something like this up to second order: $$ h_{ab}(x,y) = \frac{1}{2} \, x_a \, y_b + \frac{1}{12} \, c^{ij}{}_b \, x_i \, y_j \, \left(x_a-y_a\right)-\frac{1}{24} \, c^{lk}{}_b \,c^{ij}{}_k \, y_a x_i y_j x_l + \mathcal{O}(c^3) \,, $$ but is there a closed formula in terms of $m_i(x,y)$?