Solve $e^x+x=1$

Using the series expansion we have:

$$1+x+\frac{x^2}{2!}+ \dots + \frac{x^n}{n!}+\dots = 1-x$$

If $x$ is positive it is immediately obvious that there can be no equality.

If $x<0$ then the RHS is greater than 1 and $e^{x}<1$.

This is not strictly an "algebraic" solution, but with the term in $e^x$ we do not expect anything purely algebraic.

"Lambert W" is a hint for "algebraic solution".
The solution for $\mathrm{e}^x + x = 1$ is $1-\mathrm W(\mathrm{e})$,
to find ALL complex solutions, use all branches of the Lambert W ...

$$ \begin{align*} &\dots \\ 1 - \mathrm{W}_{-4}(\mathrm{e}) &= 3.159947300 + 23.47017395 i \\ 1 - \mathrm{W}_{-3}(\mathrm{e}) &= 2.849014724 + 17.17149358 i \\ 1 - \mathrm{W}_{-2}(\mathrm{e}) &= 2.393982241 + 10.86800606 i \\ 1 - \mathrm{W}_{-1}(\mathrm{e}) &= 1.532092122 + 4.597158013 i \\ 1 - \mathrm{W}_{0}(\mathrm{e}) &= 0.000000000 \\ 1 - \mathrm{W}_{1}(\mathrm{e}) &= 1.532092122 - 4.597158013 i \\ 1 - \mathrm{W}_{2}(\mathrm{e}) &= 2.393982241 - 10.86800606 i \\ 1 - \mathrm{W}_{3}(\mathrm{e}) &= 2.849014724 - 17.17149358 i \\ 1 - \mathrm{W}_{4}(\mathrm{e}) &= 3.159947300 - 23.47017395 i \\ 1 - \mathrm{W}_{5}(\mathrm{e}) &= 3.396557044 - 29.76478701 i \\ &\dots \end{align*} $$

explanation

$\mathrm{e}^x+x=1$
$\mathrm{e}^x=1-x$
$\mathrm{e} = (1-x)\mathrm{e}^{1-x}$
$\mathrm{W}(\mathrm{e}) = 1-x$
$x = 1-\mathrm W(\mathrm{e})$

You can see this very easily graphically. The equation is $$e^x=1-x$$ and the two sides of the equation are plotted here (from Wolfram Alpha):

The intuition for a formal proof also follows directly from the picture (the functions are both monotonic but in opposite directions), if that's your aim.