The equation
$$x\mathrm{e}^x = c$$
has a solution
$$x = W(c)$$
where $W$ is the Lambert W function and $c$ is constant.
Can the equation
$$ (x-1) \exp (x) = c $$
be solved using the Lambert W function?
Solving $(x-1)\exp x = c$ with the Lambert W Function
lambert-w
Related Solutions
$\def\H{\operatorname H}$
Series solution 1
Let $x=e^w$ and apply Lagrange reversion to get a branch of the inverse function:
$$\ln(x)-ax^2-bx=c\iff w=c+ae^{2w}+be^w\implies x=e^c+\sum_{n=1}^\infty\frac1{n!}\frac{d^{n-1}}{dc^{n-1}}e^c(ae^{2c}+be^c)^n$$
Then, use Binomial theorem:
$$\frac{d^{n-1}}{dc^{n-1}}e^c(ae^{2c}+be^c)=\sum_{m=0}^n\binom mn\frac{d^{n-1}}{dc^{n-1}}e^{c(m+n+1)}$$
Therefore:
$$\bbox[3px,border:3px solid green]{\ln(x)-ax^2-bx=c\implies x=e^c+\sum_{n=1}^\infty\sum_{m=1}^n\frac{a^m b^{n-m}e^{(m+n+1)c}}{(n-m)!m!}(m+n+1)^{n-1}}$$
Series solution 2
Apply a Tschirnhaus transformation $2ax=w-b$ and Lagrange reversion:
$$\ln(x)-ax^2-bx=c\iff w=b+2a e^{c-\frac{b^2}{4a}}e^\frac{w^2}{4a}=b+ue^{-vw^2}=b+\sum_{n=1}^\infty\frac{u^n}{n!}\frac{d^{n-1}}{db^{n-1}}e^{-vnb^2}$$
Somehow, expanding $e^x$ as a Maclaurin series, evaluating derivatives, and evaluating this sum makes it hard to find the Hermite polynomials, but recalling their Rodrigues formula:
$$\H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}$$
and $\frac{d^n f(ax)}{da^n}=a^nf^{(n)}(ax)$ gives:
$$\bbox[3px,border:3px solid blue]{w=b+ue^{-vw^2}\implies w=b-\sum_{n=1}^\infty\frac{\left(-ue^{-vb^2}\right)^n}{n!} (vn)^\frac{n-1}2\H_{n-1}(\sqrt{vn}b)}$$
This answer doesn't solve the problem but intends to simplify the ODE. $$\frac{d^2y}{dx^2} + A\frac{dy}{dx}- B\ W(Cy) = 0$$ This is a second order ODE of autonomous kind. The usual way to reduce the order is the change of function : $$\frac{dy}{dx}=u(y)\quad\implies\quad \frac{d^2y}{dx^2}=\frac{du}{dy}\frac{dy}{dx}=u\frac{du}{dy}$$ $$u(y)\frac{du}{dy} + A\:u(y)- B\ W(C\,y) = 0$$ This is a first order nonlinear ODE.
Now one has to solve for $u(y)$ an ODE of lower order. But this doesn't help us very much because this is extremely unlikely to be possible with such a special function involved. Even with simpler functions $f(y)$ solving $u(y)\frac{du}{dy} + A\:u(y)+f(y) = 0$ leads to complicated solutions or is often impossible analytically due to the non linearity. For example the Abel's differential equations of the second kind.
In addition : How to get ride of the special function.
Let $\quad W(Cy)=t(y)$
$W(Cy)e^{W(Cy)}=Cy\quad\implies\quad te^{t}=Cy\quad\implies\quad C\,dy=e^t(t+1)dt$
$$\frac{du}{dy}=\frac{du}{dt}\frac{dt}{dy}=\frac{C}{e^t(t+1)}\frac{du}{dt}$$
$$u\frac{C}{e^t(t+1)}\frac{du}{dt} + A\,u- B\,t = 0$$
$$\boxed{u\,\frac{du}{dt} + \frac{1}{C}e^t(t+1)\left(A\,u- B\,t\right) = 0}$$ This is a first order nonlinear ODE to be solved for $u(t)$.
Unfortunately even without special function into it solving this ODE for solution in terms of a finite number of standard functions seems not possible in the general case of the parameters.
Best Answer
Notice that
$$(x-1)e^x = (x-1)e^{x-1} \cdot e$$
Therefore,
$$(x-1)e^x =c \iff (x-1)e^{x-1} = \frac c e $$
Take the $W$ function of both sides:
$$x-1 = W \left( \frac c e \right)$$
Then just add $1$ and you're done!
$$x = 1+W \left( \frac c e \right)$$