Suppose $f:\mathbb{R} \to \mathbb{R}$ is an analytic, strictly increasing, surjective function that vanishes only at $x=0$.
Is there any way to compute the real root of $f(x)+c=0$ for any $c \in \mathbb{R}$ analytically? (It is obvious that $f(x)+c=0$ has only one real root).
Is it always possible to express the root of $f(x)+c=0$ in terms of elementary functions?
By your assumptions, $f(x)=-c$, $f$ is a surjective, strictly increasing, analytic function on all of $\mathbb{R}$ such that $f(0)=0 \implies x=0$.
Therefore, $f$ has a convergent infinite series everywhere and by the Lagrange inversion theorem, if $f'(-c) \neq 0$, $x=f^{-1}(-c)$ can be expressed as an infinite series:
${\displaystyle x=-c+\sum _{n=1}^{\infty }g_{n}{\frac {(z-f(-c))^{n}}{n!}},}$ where ${\displaystyle g_{n}=\lim _{w\to -c}\left[{\frac {\mathrm {d} ^{\,n-1}}{\mathrm {d} w^{\,n-1}}}\left({\frac {w+c}{f(w)-f(-c)}}\right)^{n}\right]}$
This settles down the issue of expressing the root of $f(x)+c=0$ analytically when $f'(-c) \neq 0$. The problem of whether $f^{-1}(x)$ can be expressed in terms of elementary functions is rather subtle and I have obtained partial results on this. I haven't found a counter-example yet, but I have found several examples and a class of functions for which it is possible.