Invertibility of specific function
The answer to the question of whether the inverse has a closed form depends of course on one's definition of "closed form." One plausible definition is that a closed-form function is a function that lies in a so-called Liouvillian extension of $\mathbb{C}(x)$, the field of rational functions of $x$ with complex coefficients. I won't give the exact definition of a Liouvillian extension, but suffice it to say that any function that you can get via a finite number of applications of addition, subtraction, multiplication, division, taking $n$th roots, exponentiation, and taking logarithms is going to be a closed-form function in this sense. Note that since we're working over the complex numbers, we get trig functions and their inverses as well. So this covers everything that most everyone would agree is "closed form." (Liouvillian extensions also include algebraic functions that aren't expressible using radicals; not everyone would consider such functions to be expressible in "closed form," but since we're going to show that a certain function is not expressible in closed form, it doesn't hurt to include extra functions in our class of "closed-form functions.")
Rigorous proofs that specific functions of interest are not Liouvillian go back, naturally, to Liouville, with later contributions by other authors (e.g., Ritt, as mentioned by Iosif Pinelis). Again, reviewing the general theory is beyond the scope of a MathOverflow answer, but fortunately, when it comes to finding inverses, there is a theorem of Rosenlicht ("On the explicit solvability of certain transcendental equations," Publications Mathématiques de l'IHÉS 36 (1969), 15–22) that can be used to handle many of the simple "transcendental equations" that arise in practice. Stated slightly informally, the relevant special case of Rosenlicht's theorem is the following.
Theorem. Suppose that $y_1, \ldots, y_n$ and $z_1, \ldots, z_n$ are closed-form functions of $x$ satisfying $$\frac{y_i'}{y_i} = z_i', \qquad i=1,\ldots,n.$$ If $\mathbb{C}(x,y_1, \ldots,y_n,z_1,\ldots,z_n)$ is algebraic over both $\mathbb{C}(x,y_1,\ldots,y_n)$ and $\mathbb{C}(x,z_1,\ldots,z_n)$, then $y_1,\ldots,y_n$ and $z_1,\ldots z_n$ are all algebraic over $\mathbb{C}(x)$.
In light of Pietro Majer's observation, let's use this theorem to show that the function $f$ defined implicitly by the equation $x = f - \sin f$ (Kepler's equation, as noted by Rob Corless) has no closed-form expression. We're only going to need the special case $n=1$ of the theorem. The first step is to write everything in terms of exponentials. Recall that $\sin f = (e^{if}-e^{-if})/2i$, so if we set $z := if$ then we have the equation $$x = -iz - \frac{e^z - e^{-z}}{2i}.\qquad\qquad(*)$$ The equation $y'\!/y = z'$ appearing in Rosenlicht's theorem is secretly the equation $y=e^z$ in disguise. So what we need to do is to introduce extra functions to represent the exponentials that appear, to turn our equations into polynomial equations. Here, all we need to do is to set $y=e^z$. Then $y'\!/y = z'$ and Equation $(*)$ becomes $$x = -iz- \frac{y - 1/y}{2i}.\qquad\qquad (**)$$ We're now ready to apply Rosenlicht's theorem with $n=1$. Certainly $\mathbb{C}(x,y,z)$ is algebraic over $\mathbb{C}(x,y)$ because $z$ is actually a rational function of $x$ and $y$. It's also true that $\mathbb{C}(x,y,z)$ is algebraic over $\mathbb{C}(x,z)$ because $y$ satisfies a quadratic equation with coefficients that are polynomial (in fact, linear) in $x$ and $z$. The hypothesis of the theorem is therefore satisfied. What does this tell us?
Well, if $f$ is a closed-form function of $x$, then so is $z=if$ as well as $y=e^z = e^{if}$. So if $f$ is a closed-form function, then Rosenlicht's theorem tells us that $y$ and $z$ must in fact be algebraic functions of $x$.
This isn't quite a contradiction yet, but it's not so hard to show that $y$ and $z$ can't be algebraic functions of $x$. We can use the argument given by Bronstein et al. in their paper showing that the Lambert $W$ function is not Liouvillian ("Algebraic properties of the Lambert $W$ function from a result of Rosenlicht and Liouville," Integral Transforms and Special Functions 19 (2008), 709–712). If $z$ has a pole of finite order (in the extended complex plane), then $y$ has an essential singularity, but this contradicts Equation $(**)$ since the left-hand side has no singularities. So $z$ must be constant, but this is absurd.
By the way, according to Rosenlicht, Liouville himself already knew that the solution to Kepler's equation is not Liouvillian, but I haven't checked Liouville's paper myself.
[I'm making this answer community wiki since I benefited from the observations of several other respondents.]
The change of variable given above by Pietro Majer shows that this is equivalent to Kepler's equation Wikepedia on Kepler's Equation which is believed not to have any closed form solution (let alone an elementary solution). I am actually not so sure that's true and I don't know of any proof.
According to Ritt's Theorem 23 (page 89) (see also the first paragraph on page 90 of that paper and the definition of (Liouville) monomials of the first order on page 70 there), your desired inverse is an elementary function only if $t-\sin t\,\cos t$ is an algebraic function of $e^{v(t)}$ or $\ln v(t)$ for some algebraic function $v$. However, at this point I do not see how to check the latter condition in your case.
Mathematica cannot invert your function either:
