Inverse function of $y=W(e^{ax+b})-W(e^{cx+d})+zx$

I have a simple question for which I am looking for a closed form expression (If there exits one). In other words, given:

$$y=W(e^{ax+b})-W(e^{cx+d})+zx$$

where $W$ is the Lambert $W$ function and $a,b,c,d,z$ are some constants, what is the function $f$, such that

$$x=f(y)$$

Thanks alot in advance.

EDIT : If there exists no closed form solution, I will be happy to see nice arguments supporting this.

$\rightarrow$EDIT2 : As can be seen, we have a solution for the simplified version of this problem. If there exists a solution I have the following ideas to resolve the full version of the problem:

$1$- Is it possible to write

$$W(e^{f(x)})=W(e^{ax+b})-W(e^{cx+d})$$

$2$- Having

$$y_1=W(e^{ax+b})+z_1x$$ and $$y_2=-W(e^{cx+d})+z_2x$$

where $z=z_1+z_2$, $y=y_1+y_2$ and $f^{-1}(y_1)$ and $f^{-1}(y_2)$ are known functions as already found. Can we say that $f^{-1}(y)=f^{-1}(y_1)+f^{-1}(y_2)$? or can we modify this idea to get somethign useful?

Answer

$y=W(e^{ax+b})-W(e^{cx+d})+zx$

$W(e^{cx+d})+y-zx=W(e^{ax+b})$

$(W(e^{cx+d})+y-zx)e^{W(e^{cx+d})+y-zx}=e^{ax+b}$

$(W(e^{cx+d})+y-zx)e^{W(e^{cx+d})}e^{y-zx}=e^{ax+b}$

$(W(e^{cx+d})+y-zx)\dfrac{e^{cx+d}}{W(e^{cx+d})}=e^{(a+z)x+b-y}$

$W(e^{cx+d})+y-zx=e^{(a-c+z)x+b-d-y}W(e^{cx+d})$

$(e^{(a-c+z)x+b-d-y}-1)W(e^{cx+d})=y-zx$

$W(e^{cx+d})=\dfrac{y-zx}{e^{(a-c+z)x+b-d-y}-1}$

$e^{cx+d}=\left(\dfrac{y-zx}{e^{(a-c+z)x+b-d-y}-1}\right)e^{\frac{y-zx}{e^{(a-c+z)x+b-d-y}-1}}$

Then you can only force to use Lagrange inversion theorem or Lagrange reversion theorem unless the special cases when $a=0$ or $c=0$ or $a-c+z=0$ .