Since $f(x)>0$ for all $x>0$, it suffices to prove the inequality
$$y+f(y)\geq 2$$
for all $y>0$. In other words, we need to show
$$y^{-e^{y-1}}\geq 2-y.$$
This is clear for $y\geq 2$. Otherwise, taking logarithms, we need to show
$$-e^{y-1}\ln(y)\geq \ln(2-y)$$
for $0<y<2$. Now, substitute $y=1+t$ for $-1<t<1$ to get that we need
$$-e^t\ln(1+t)\geq \ln(1-t)\Leftrightarrow e^{t/2}\ln(1+t)+e^{-t/2}\ln(1-t)\leq 0.$$
Writing
$$e^{t/2}\ln(1+t)=\sum_{k=0}^\infty a_kt^k,$$
we need
$$\sum_{\ell=0}^\infty 2a_{2\ell}t^{2\ell}\leq 0,$$
so it suffices to show that $a_{2\ell}\leq 0$ for all integers $\ell\geq 0$. (The reason the original inequality is so sharp, especially in the neighborhood of $y=1$, is because $a_0=a_2=0$.) Using the Taylor series expansions for $e^x$ and $\ln(1+x)$, we can write
$$-a_{2\ell}=\sum_{k=0}^{2\ell-1}\frac{(-1/2)^k}{k!(2\ell-k)}=\sum_{j=0}^{\ell-1}\left(\frac{1}{2^{2j}(2j)!(2\ell-2j)}-\frac1{2^{2j+1}(2j+1)!(2\ell-2j-1)}\right).$$
So, it will be enough to show
$$\frac{1}{2^{2j}(2j)!(2\ell-2j)}\geq \frac1{2^{2j+1}(2j+1)!(2\ell-2j-1)}$$
for $0\leq j<\ell$. Cancelling like terms, this is equivalent to
$$(2\ell-2j-1)(2j+1)\geq \ell-j,$$
which holds since
$$(2\ell-2j-1)(2j+1)\geq 2\ell-2j-1=(\ell-j)+(\ell-j-1)\geq \ell-j.$$
This finishes the proof.
Note that your generalization to sums of repeated applications of $f$ follows from applying the inequality $x+f(x)\geq 2$ at various values of $x$, while your intermediate conjecture $x\leq f(f(x))$ is false for $x<1$.
Remark: Although we get a closed-form bound (in terms of elementary functions), the solution is still quite complicated.
We have
$$\int_{-\infty}^0 (\cosh x - 1)^x\, \mathrm{d}x
= \int_0^\infty 2^{-x}\left(\sinh\frac{x}{2}\right)^{-2x}\,\mathrm{d} x = \int_0^\infty 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x.$$
Fact 1: $\int (\sinh x)^{-5}\, \mathrm{d}x =
-\frac{\cosh x}{4\sinh^4 x} + \frac{3\cosh x}{8\sinh^2 x} - \frac{3}{8}\ln\frac{\mathrm{e}^x + 1}{\mathrm{e}^x - 1} + C$.
Fact 2: $\int (\sinh x)^{-4}\, \mathrm{d}x =
-\frac{\cosh x}{3\sinh^3 x} + \frac{2\cosh x}{3\sinh x} + C$.
Fact 3: $\sinh x \ge x + \frac16 x^3$ for all $x\in [0, 1]$.
Let $T(f(x),a,n)$ denote the $n$-th order Taylor approximation of $f(x)$ at $x = a$.
Fact 4: For all $x\in [0, 1]$,
$(1 + x^2/6)^{-4x} \le T((1 + x^2/6)^{-4x}, 0, 12)$.
(Proof: Take logarithm, then take derivative and the 2nd derivative.)
Fact 5: For all $x\in [0, 1]$,
$2^{-2x} \le T(2^{-2x}, 0, 6)$.
(Proof: Take logarithm, then take derivative.)
Fact 6: For all $u \in [-1, 0]$,
$\mathrm{e}^{-4u} \le T(\mathrm{e}^{-4u}, 0, 8) + \frac{512}{315}u^8$.
(Proof: Take logarithm, then take derivative.)
First, using Fact 1, we have
\begin{align*}
\int_{5/4}^\infty 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x
&\le \int_{5/4}^\infty 2^{1 - 2\cdot \frac{5}{4}}(\sinh x)^{-4\cdot \frac{5}{4}}\, \mathrm{d}x \\
&= -\frac{\sqrt2}{4}\left(-\frac{\cosh (5/4)}{4\sinh^4 (5/4)} + \frac{3\cosh (5/4)}{8\sinh^2 (5/4)} - \frac{3}{8}\ln\frac{\mathrm{e}^{5/4} + 1}{\mathrm{e}^{5/4} - 1}\right)\\
&< \frac{3}{500}.
\end{align*}
Second, using Fact 2, we have
\begin{align*}
\int_1^{5/4} 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x
&\le \int_1^{5/4} 2^{1 - 2\cdot 1}(\sinh x)^{-4\cdot 1}\, \mathrm{d}x\\
&= -\frac{\cosh (5/4)}{6\sinh^3 (5/4)} + \frac{\cosh (5/4)}{3\sinh (5/4)} - \left(-\frac{\cosh 1}{6\sinh^3 1} + \frac{\cosh 1}{3\sinh 1}\right)\\
&< \frac{3}{80}.
\end{align*}
Third, using Facts 3-6, we have
\begin{align*}
&\int_0^1 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x\\
\le\,& \int_0^1 2^{1 - 2x}(x + x^3/6)^{-4x}\, \mathrm{d}x\\
=\, & \int_0^1 2^{1 - 2x}(1 + x^2/6)^{-4x}x^{-4x}\, \mathrm{d}x\\
\le\, & 2\int_0^1 T(2^{-2x}, 0, 6) \, T((1 + x^2/6)^{-4x}, 0, 12)
\, \left[T(\mathrm{e}^{-4u}, 0, 8) + \frac{512}{315}u^8\right]_{u = x\ln x}\, \mathrm{d} x \tag{1}\\
<\, & 1549/500.
\end{align*}
Note: The integral in (1) admits a closed form
$a_0 + a_1\ln 2 + a_2\ln^2 2 + a_3\ln^3 2 + a_4\ln^4 2 + a_5\ln^5 2 + a_6\ln^6 2$
where $a_i$'s are all rational numbers.
Thus, we have $\int_0^\infty 2^{1 - 2x}(\sinh x)^{-4x}\, \mathrm{d}x
< \frac{3}{500} + \frac{3}{80} + 1549/500 < \pi$.
We are done.
Best Answer
You may use the well-known inequality $e^x\ge 1+x$, therefore:
$$x^\frac{\sin(x)}{x}=e^{\ln(x)\frac{\sin(x)}{x}}\ge 1+\ln(x)\frac{\sin(x)}{x}$$
So it suffices to prove that:
$$1+\ln(x)\frac{\sin(x)}{x}>\sin(x)+\frac{1}{x-1}$$
Or:
$$\sin(x)\left(1-\frac{\ln(x)}{x}\right)<1-\frac{1}{x-1}$$
Now if $\pi\le x\le2\pi$ you have LHS$\,\le 0$ and RHS$\,>0$ and you are done.
If $x>2\pi$ you may ignore $\sin(x)$ and directly prove that:
$$1-\frac{\ln(x)}{x}<1-\frac{1}{x-1}$$
Or:
$$\ln(x)>\frac{x}{x-1}$$
The last inequality is obvious ($\ln$ is increasing, $\frac{x}{x-1}$ is decreasing, so $\ln(x)>\ln(2\pi)>\frac{2\pi}{2\pi-1}>\frac{x}{x-1}$)
Note that $\ln(x)>\frac{x}{x-1}$ is not true for $1\le x\le 3.85\ldots$, that's why $\pi\le x\le 2\pi$ had to be proved separately.