Closed-form of $\int_0^\infty \frac{1}{\left(a+\cosh x\right)^{1/n}} \, dx$ for $a=0,1$

For the first one, $$\begin{align} \int_0^{\infty} (\operatorname{sech}x)^{2s}dx \\ &= \int_0^{\infty} (\operatorname{sech}^2x)^{s-1}\operatorname{sech}^2x\, dx \\ &= \int_0^{\infty} (1-\tanh^2x)^{s-1}\,\mathrm{d}(\tanh x)\\ &= \int_0^1 (1-x^2)^{s-1} \mathrm{d}x\\ &= \frac12 \int_0^1 (1-x)^{s-1} x^{-\frac12} \mathrm{d}x\\ &= \frac12 B(s,\frac12)=\frac{\sqrt{\pi}}{2}\frac{\Gamma(s)}{\Gamma(\frac12+s)} \end{align}$$ and so your conjecture is correct. For the second integral, let $x=2t$: $$\begin{align} \int_0^{\infty} \frac{1}{(1+\cosh x)^s}\mathrm{d}x \\ &= 2\int_0^{\infty} \frac{1}{(1+\cosh 2t)^s}\mathrm{d}t \\ &= 2\int_0^{\infty} \frac{1}{(2\cosh^2(t))^s}\mathrm{d}t\\ &= 2^{1-s} \int_0^{\infty} (\operatorname{sech}t)^{2s}\mathrm{d}t\\ &= 2^{-s} B(s,\frac12)=\frac{\sqrt{\pi}}{2^s}\frac{\Gamma(s)}{\Gamma(\frac12+s)} \end{align}$$

Extending @nospoon's idea, we notice that

$$ a + \cosh 2x = (a+1)\cosh^2 x (1 - b \tanh^2 x), \qquad b =\frac{a-1}{a+1}. $$

If $a > -1$, then $b < 1$ and using the substitution $u = \tanh^2 x$ we get

$$ \int_{0}^{\infty} \frac{dx}{(a + \cosh x)^{s}} \, dx = \frac{1}{(a+1)^{s}} \int_{0}^{1} \frac{(1 - u)^{s-1}}{(1 - b u)^s\sqrt{u}} \, du. $$

Making further substitution $v = \frac{1-u}{1-bu}$, we have

$$ \int_{0}^{\infty} \frac{dx}{(a + \cosh x)^{s}} \, dx = \frac{1}{(a+1)^{s}} \int_{0}^{1} \frac{v^{s-1}}{\sqrt{(1-v)(1-bv)}} \, dv. $$

This is easily integrated when $b = 0$ or $b = -1$, each correspondingto $a = 1$ or $a = 0$, but I doubt that this integral has a nice closed form in general. For example, when $s = 1/2$, this becomes an elliptic integral and Mathematica 11 gives

$$ \int_{0}^{1} \frac{dv}{\sqrt{v(1-v)(1-bv)}} = 2 \left[ \frac{1}{\sqrt{b}} K \left( \sqrt{\tfrac{1}{b}} \right) + i K \left( \sqrt{1-b} \right) \right], \quad 0 < b < 1, $$

where $K(k)$ is the complete elliptic integral of the 1st kind.