The result below assumes $t>0$ (as usual for Laplace transforms).
In the Bromwich contour $\gamma$ has to be chosen large enough that it is to the right of all singularities (poles and branch points) so $\gamma = 0^+$ is perfectly valid. The singularities of $\log s/(1+s)$ are a pole at $s=-1$ and two branch points at $s=0$ and $s=\infty$ which we connect via a branch cut along the negative real line.
Then we can deform the contour further to a path which starts at $-\infty -i 0^+$. Runs along the negative real line just below the branch cut. Ends at $0-i 0^+$ in a little semi-circle and then runs back from $0+i 0^+$ to $-\infty +i0^+$ just above the branch cut.
The Bromwich integral thus is given by
$$f(t)=\frac1{2\pi i} \int_{-\infty}^0\!dx\, \left(
\frac{\log (x-i0^+)}{1+x-i0^+ } - \frac{\log (x+i0^+)}{1+x+i0^+ } \right) e^{x t} $$
as the small circle around the branch point at $0$ does not contribute ($|z|\log z \to 0$ for $|z|\to0$).
In the remaining integral, we use $\log(x \pm i 0^+) = \log |x| \pm i \pi$ valid for $x<0$:
$$\begin{align} f(t) &= \frac1{2\pi i} \int_{-\infty}^0\!dx\, \left(
\frac{\log |x|-i\pi}{1+x-i0^+ } - \frac{\log |x|+i\pi}{1+x+i0^+} \right) e^{x t}\\
&= \frac1{2\pi i} \overbrace{\int_{-\infty}^0\!dx\, \log |x| \underbrace{\left(
\frac1{1+x-i0^+ } - \frac1{1+x+i0^+}\right)}_{2\pi i \delta(x+1)} e^{x t}}^{=0}\\
&\quad -\frac12 \int_{-\infty}^0\!dx \underbrace{\left(
\frac1{1+x-i0^+ } + \frac1{1+x+i0^+}\right)}_{2\mathcal{P}\,(1+x)^{-1}} e^{x t} \\
&= -\int_{-\infty}^0\!dx \,\mathcal{P} \frac{e^{x t}}{1+x}
=- e^{-t} \int_{-\infty}^t\!ds \,\mathcal{P} \frac{e^{s}}{s}\\
&=- e^{-t} \mathop{\rm Ei}(t)
\end{align}$$
with $s=(1+x)t$ and Ei the exponential integral.
I will illustrate a slightly different, more rigorous derivation of the ILT result here, which is
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{-\sqrt{s (1+s)}} \, e^{s t} = \frac12 e^{-t/2} \frac{I_1 \left ( \frac12 \sqrt{t^2-1} \right )}{\sqrt{t^2-1}} H(t-1) $$
where
$$H(t-1) = \begin{cases} 1 \quad t \gt 1 \\ 0 \quad t \le 1 \end{cases} $$
To demonstrate where this result comes from, I am going to define the following contour integral:
$$ \oint_C dz \, e^{-\sqrt{z (1+z)}} \, e^{t z} $$
where we will figure out the allowed values of $t$ and the contour $C$ is as follows:
where the radius of the large arc is $R$ and the radii of the small arcs are each $\epsilon$. By Cauchy's theorem, this contour integral is zero. On the other hand, we can write out the contour integral in the limit as $\epsilon \to 0$ as follows:
$$ \int_{c-i \sqrt{R^2-c^2}}^{c+i \sqrt{R^2-c^2}} ds \, e^{-\sqrt{s (1+s)}} \, e^{s t} + i R \int_{\pi/2 - \arcsin{(c/R)}}^{\pi} d\theta \, e^{i \theta} \, e^{-R e^{i \theta} \left (1+\frac1{R e^{i \theta}} \right )^{1/2}} e^{R t e^{i \theta}} \\+ e^{i \pi} \int_R^{1} dx \, e^{\sqrt{x (x-1)}} e^{-x t} + e^{i \pi} \int_1^0 dx \, e^{-i \sqrt{x (1-x)}} e^{-x t} \\ + e^{-i \pi} \int_0^1 dx \, e^{i \sqrt{x (1-x)}} e^{-x t} + e^{-i \pi} \int_1^R dx \, e^{\sqrt{x (x-1)}} e^{-x t} \\ + i R \int_{\pi}^{3 \pi/2 + \arcsin{(c/R)}} d\theta \, e^{i \theta} \, e^{-R e^{i \theta} \left (1+\frac1{R e^{i \theta}} \right )^{1/2}} e^{R t e^{i \theta}}$$
It should be clear that the third and sixth integrals cancel. Further, we can deduce the values of $t$ for which the ILT is defined by combining the second and seventh integrals:
$$\begin{align} i R \int_{\pi/2-\arcsin{(c/R)}}^{3 \pi/2+\arcsin{(c/R)}} d\theta \, e^{i \theta} \, e^{-R e^{i \theta} \left (1+\frac1{R e^{i \theta}} \right )^{1/2}} e^{R t e^{i \theta}} &= R \int_{\frac{c}{R} + i \sqrt{1-\frac{c^2}{R^2}}}^{\frac{c}{R} - i \sqrt{1-\frac{c^2}{R^2}}} d\zeta \, e^{-R \zeta \left ( 1 + \frac1{R \zeta} \right)^{1/2}} e^{R t \zeta} \\ &=_{R \to \infty} R e^{-1/2} \int_{\frac{c}{R}+i}^{\frac{c}{R}-i} d\zeta \, e^{R (t-1) \zeta} \\ &= -i 2 e^{c (t-1)} e^{-1/2} \frac{\sin{R (t-1)}}{t-1} \\ &=_{R \to \infty} -i 2 \pi e^{c (t-1)} e^{-1/2} \delta(t-1) \end{align} $$
It should be noted that the above integrals converge as $R \to \infty$ only when $t \gt 1$. Accordingly, there will be a $H(t-1)$ term in the final result. Because $\lim_{x \to 0} H(x) \delta(x) = 0$, we may ignore the delta function contribution in this case. (When $t \lt 1$, the ILT is zero.)
Now, as $R \to \infty$ we are down to three integrals which we can relate by Cauchy's theorem:
$$ \int_{c-i \infty}^{c+i \infty} ds \, e^{-\sqrt{s (1+s)}} \, e^{s t} = \int_0^1 dx \, e^{i \sqrt{x (1-x)}} e^{-x t} - \int_0^1 dx \, e^{-i \sqrt{x (1-x)}} e^{-x t}$$
so that the ILT for $t \gt 1$ is now in terms of a single, real integral:
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{-\sqrt{s (1+s)}} \, e^{s t} = \frac1{\pi} \int_0^1 dx \, \sin{\left ( \sqrt{x (1-x)} \right )} e^{-t x} H(t-1)$$
We may now focus on evaluating this real integral. We can put the integral into a more familiar form with the substitution $x=\sin^2{(\theta/2)}$:
$$\begin{align}\frac1{\pi} \int_0^1 dx \, \sin{\left ( \sqrt{x (1-x)} \right )} e^{-t x} &= \frac1{2 \pi} e^{-t/2} \int_0^{\pi} d\theta \, \sin{\theta} \, \sin{\left ( \frac12 \sin{\theta} \right )} e^{(t/2) \cos{\theta}} \\ &= \frac1{4 \pi} e^{-t/2} \operatorname{Im} \int_{-\pi}^{\pi} d\theta \, \sin{\theta} \, e^{i \frac12 \sin{\theta}} e^{(t/2) \cos{\theta}} \end{align} $$
We may evaluate the integral on the right by expressing it as a complex integral and taking it from there...
$$\begin{align} \int_{-\pi}^{\pi} d\theta \, \sin{\theta} \, e^{i \frac12 \sin{\theta}} e^{(t/2) \cos{\theta}} &= -i \oint_{|z|=1} \frac{dz}{z} \left ( \frac{z-z^{-1}}{i 2} \right ) e^{(t/2) \left (\frac{z+z^{-1}}{2} \right )} e^{(i/2) \left (\frac{z-z^{-1}}{i 2} \right )} \\ &= -\frac12 \oint_{|z|=1} \frac{dz}{z} \left ( z-z^{-1} \right ) e^{\frac14 (t+1) z + \frac14 (t-1) z^{-1}} \end{align} $$
The contour integral is equal to $i 2 \pi$ times the sum of the residues of the poles inside of the unit circle. In this case, the integrand has an essential singularity at $z=0$. To compute the residue at $z=0$, we expand the exponential in a Laurent series and we need only determine the coefficient of $z^{-1}$ in that expansion. In short, we wish to compute the coefficient of $z^0$ in the Laurent expansion of
$$\left ( z-z^{-1} \right ) e^{a z + b z^{-1}} $$
where $a=\frac14 (t+1)$ and $b=\frac14 (t-1)$. This expansion looks like
$$\begin{align} \left ( z-z^{-1} \right ) e^{a z + b z^{-1}} &= \left ( z-z^{-1} \right ) \sum_{k=0}^{\infty} \frac1{k!} \left ( a z + b z^{-1} \right )^k \\ &= \left ( z-z^{-1} \right ) \sum_{k=0}^{\infty} \frac1{k!} \sum_{m=0}^k \binom{k}{m} (a z)^m \left ( b z^{-1} \right )^{k-m} \\ &= \left ( z-z^{-1} \right ) \sum_{k=0}^{\infty} b^k \sum_{m=0}^k \frac1{m! (k-m)!} \left ( \frac{a}{b} \right )^m z^{2 m-k} \end{align} $$
To simplify things a bit, we note that only odd values of $k$ will result in a nonzero coefficient of $z^0$. We can then rewrite that last sum as
$$ \left ( z-z^{-1} \right ) e^{a z + b z^{-1}} = \left ( z-z^{-1} \right ) \sum_{k=0}^{\infty} b^{2 k+1} \sum_{m=0}^k \frac1{m! (2 k+1-m)!} \left ( \frac{a}{b} \right )^m z^{2 m-2 k - 1} $$
We have nonzero coefficients of $z^0$ when $m=k$ or $m=k+1$. Therefore, by the residue theorem, the integral above is equal to
$$\int_{-\pi}^{\pi} d\theta \, \sin{\theta} \, e^{i \frac12 \sin{\theta}} e^{(t/2) \cos{\theta}} = -\frac12 i 2 \pi (b-a) \sum_{k=0}^{\infty} \frac{(a b)^k}{k! (k+1)!} = -i \pi \frac{b-a}{\sqrt{a b}} I_1 \left ( 2 \sqrt{a b} \right )$$
Plug in $a=\frac14 (t+1)$ and $b=\frac14 (t-1)$ and we find that
$$\frac1{\pi} \int_0^1 dx \, \sin{\left ( \sqrt{x (1-x)} \right )} e^{-t x} = \frac12 e^{-t/2} \frac{I_1 \left ( \frac12 \sqrt{t^2-1} \right )}{\sqrt{t^2-1}} $$
and the ILT is as asserted above.
Best Answer
Consider the contour integral
$$\oint_C dz \frac{e^{t z}}{\sqrt{z^2-a^2}} $$
where $C$ is the contour drawn above, and $t \gt 0$. By Cauchy's theorem, this integral is zero. However, to evaluate the ILT, we need to evaluate all of the pieces of the contour integral. Thankfully, the OP has provided a diagram with such nice labels. Thus,
$$\int_{AB} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = \int_{\beta-i R}^{\beta+i R} ds \frac{e^{t s}}{\sqrt{s^2-a^2}}$$
$$\int_{BC} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = i R \int_{\pi/2}^{\pi} d\theta \, e^{i \theta} \frac{e^{t R e^{i \theta}}}{\sqrt{R^2 e^{i 2 \theta}-a^2}} $$
$$\int_{CD} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = \int_{-R}^{-a-i \epsilon} dx \frac{e^{t x}}{\sqrt{x^2-a^2}} $$
$$\int_{DE} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{t(-a+\epsilon e^{i \phi})}}{\sqrt{(-a+\epsilon e^{i \phi})^2-a^2}}$$
$$\int_{EF} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = \int_{-a+\epsilon}^{a-\epsilon} dx \frac{e^{t x}}{e^{i \pi/2} \sqrt{a^2-x^2}} $$
$$\int_{FG} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac{e^{t(a+\epsilon e^{i \phi})}}{\sqrt{(a+\epsilon e^{i \phi})^2-a^2}}$$
$$\int_{GH} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = \int_{a+\epsilon}^{-a-\epsilon} dx \frac{e^{t x}}{e^{-i \pi/2} \sqrt{a^2-x^2}} $$
$$\int_{HI} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{e^{t(-a+\epsilon e^{i \phi})}}{\sqrt{(-a+\epsilon e^{i \phi})^2-a^2}}$$
$$\int_{IJ} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = \int_{-a-i \epsilon}^{-R} dx \frac{e^{t x}}{\sqrt{x^2-a^2}} $$
$$\int_{JA} dz \frac{e^{t z}}{\sqrt{z^2-a^2}} = i R \int_{\pi}^{3 \pi/2} d\theta \, e^{i \theta} \frac{e^{t R e^{i \theta}}}{\sqrt{R^2 e^{i 2 \theta}-a^2}} $$
OK, there's a lot there, but it's not nearly as bad as it looks. The integral over $AB$ will be $i 2 \pi$ times the ILT as $R \to \infty$. The integral over $BC$ vanishes in this limit because its magnitude is bounded by
$$\frac{R}{\sqrt{R^2-a^2}} \int_0^{\pi/2} d\theta \, e^{-t R \sin{\theta}} \le \frac{R}{\sqrt{R^2-a^2}} \int_0^{\pi/2} d\theta \, e^{-2 t R \theta/\pi} \le \frac{\pi}{2 t \sqrt{R^2-a^2}}$$
The integral over $JA$ vanishes for similar reasons. The integrals over $CD$ and $IJ$ cancel each other out. The integrals over $DE$, $HI$, and $FG$ vanish as $\epsilon \to 0$. Thus, in these limits, we may write the ILT as follows:
$$\int_{\beta-i \infty}^{\beta+i \infty} ds \frac{e^{t s}}{\sqrt{s^2-a^2}} - i 2 \int_{-a}^a dx \frac{e^{t x}}{\sqrt{a^2-x^2}} = 0$$
or
$$\frac1{i 2 \pi} \int_{\beta-i \infty}^{\beta+i \infty} ds \frac{e^{t s}}{\sqrt{s^2-a^2}} = \frac1{\pi} \int_{-a}^a dx \frac{e^{t x}}{\sqrt{a^2-x^2}} $$
We may evaluate the integral on the RHS as follows. Sub $x=a \cos{u}$; then the integral is equal to
$$\frac1{\pi} \int_0^{\pi} du \, e^{a t \cos{u}} = I_0(a t)$$
where $I_0$ is the modified Bessel function of the first kind of zeroth order. Thus,