While the common way to derive it is by using the Lagrange Inverse Theorem, there technically isn't anything stopping us from making a Taylor Series for it as you would with any other function. As always, we're going to need a list of derivatives. The first one can be found pretty easily via implicit differentiation as follows:
$$y = W(x)$$
$$ye^y = x$$
$$\frac{d}{dx}\left(ye^y=x\right)$$
$$\left(y+1\right)e^y\cdot\frac{dy}{dx}=1$$
$$\therefore \frac{d}{dx}W\left(x\right)=\frac{e^{-W\left(x\right)}}{\left(W\left(x\right)+1\right)}$$
$$\frac{d}{dx}W\left(x\right)=\frac{W\left(x\right)}{x\left(W\left(x\right)+1\right)}$$
Also, to anyone unfamiliar with the last step, it is one of the main identities of the Lambert W function.
Now that we have the first derivative, we can simply differentiate as many times as we want to get all subsequent derivatives. The important thing to note, however, is that all subsequent derivatives will only require the W function to be evaluated at x. Therefore, if we know the value of W(x) we can theoretically calculate the value of any nth derivative of W(x) at that x. We can use this to our advantage by considering a value of W which is easy to calculate, such as W(e). This can be calculated as follows:
$$y=W\left(e\right)$$
$$ye^y=e=1e^1$$
$$\therefore y=1 \Rightarrow W(e)=1 $$
We now have everything we need to calculate a Taylor Series centered at x=e as usual.
$$\sum_{n=0}^{\infty}\frac{W^{\left(n\right)}\left(e\right)}{n!}\left(x-e\right)^n$$
$$=1+\frac{1\left(x-e\right)}{2\cdot e\cdot1!}-\frac{3\left(x-e\right)^2}{2^3e^2\cdot2!}+\frac{19\left(x-e\right)^3}{2^5e^3\cdot3!}-\frac{185\left(x-e\right)^4}{2^7e^4\cdot4!}+\frac{2437\left(x-e\right)^5}{2^9e^5\cdot5!}...$$
I agree that this isn't as useful as the Lagrange Inverse method (mostly due to the lack of an explicit definition for the coefficients), however I felt it was worth noting as the question did ask if there were any other methods. And, as shown by the plot of the 5th degree polynomial, this does
indeed work.
5th Degree Polynomial Approximation of W(x)
My solution (notation: we are looking for $x$ expressed with $y$ which was the previous left hand side):
Write $x=e^t$:
$$y=Ae^t+Bt e^t=e^t(A+Bt)$$
write $A/B+t=u$ to get rid of addition (you want a standard form of $xe^x$ to use the Lambert's function):
$$y=Bu e^{u-A/B}=Be^{-A/B}u e^u$$
Divide to get to the standard form:
$$e^{A/B}\frac{y}{B}=ue^u$$
Invert:
$$u=W\left(e^{A/B}\frac{y}{B}\right)$$
Now just express $x=e^{t}=e^{u-A/B}$:
$$x=e^{-A/B}\exp\left[W\left(e^{A/B}\frac{y}{B}\right)\right]$$
It's not exactly the same form as you expected, but you can use other identities, such as
$$e^{W(x)}=\frac{x}{W(x)}$$ to transform it into
$$x=\frac{y}{BW\left(e^{A/B}\frac{y}{B}\right)}$$
In your expression, $A$ and $y$ (or actually $x$) are reversed, I'm not sure about that.
Best Answer
One has $W(e^y)e^{W(e^y)} = e^y$ by definition of the Lambert function, hence $W(e^y) = e^{y-W(e^y)}$ and finally $x = \ln W(e^y) = y - W(e^y)$.