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.
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,
$$\frac1{i 2 \pi} \int_{\beta-i \infty}^{\beta+i \infty} ds \frac{e^{t s}}{\sqrt{s^2-a^2}} = I_0(a t)$$
Best Answer
The best I can do is reduce the expression for the ILT to a real integral. Based on your diagram, I get the following expression for the contour integral about the Bromwich contour:
$$\int_{c-i \infty}^{c+ i \infty} ds \, \frac{e^{s t}}{s^{3/2} \sqrt{1+ a b \frac{\tanh{\sqrt{a s}}}{\sqrt{a s}}}} + e^{i \pi} \int_{\epsilon}^R dx \, \frac{e^{-x t}}{e^{i 3 \pi/2} x^{3/2} \sqrt{1+ a b \frac{\tan{\sqrt{a x}}}{\sqrt{a x}}}} \\ + e^{-i \pi} \int_R^{\epsilon} dx \, \frac{e^{-x t}}{e^{-i 3 \pi/2} x^{3/2} \sqrt{1+ a b \frac{\tan{\sqrt{a x}}}{\sqrt{a x}}}} + i \epsilon^{-1/2} \int_{\pi}^{-\pi} d\phi \, e^{-i \phi/2} \frac{e^{\epsilon e^{i \phi} t}}{\sqrt{1+ a b \frac{\tanh{\sqrt{a \epsilon e^{i \phi}}}}{\sqrt{a \epsilon e^{i \phi}}}}} $$
A few observations. Note how I rewrote the integrand of the ILT. The function inside the square root in the denominator is now an even function of $s$, so that there are no further branch points except the factor of $s^{3/2}$ in the denominator. Also note that the integrals appear singular in $\epsilon$; I will show that the singularities cancel, as they so often do in ILT computations.
By Cauchy's theorem, the contour integral is zero. Thus, we have as $R \to \infty$:
$$\int_{c-i \infty}^{c+ i \infty} ds \, \frac{e^{s t}}{s^{3/2} \sqrt{1+ a b \frac{\tanh{\sqrt{a s}}}{\sqrt{a s}}}} = -i 2 \int_{\epsilon}^{\infty} dx \, x^{-3/2} \frac{e^{-x t}}{\sqrt{1+ a b \frac{\tan{\sqrt{a x}}}{\sqrt{a x}}}} \\+ i \epsilon^{-1/2} \int_{-\pi}^{\pi} d\phi \, e^{-i \phi/2} \frac{e^{\epsilon e^{i \phi} t}}{\sqrt{1+ a b \frac{\tanh{\sqrt{a \epsilon e^{i \phi}}}}{\sqrt{a \epsilon e^{i \phi}}}}} $$
To see how the singularities in $\epsilon$ cancel, rewrite the RHS as
$$-i 2 \int_{\epsilon}^{\infty} dx \, x^{-3/2} \frac{e^{-x t}-1}{\sqrt{1+ a b \frac{\tan{\sqrt{a x}}}{\sqrt{a x}}}} -i 2 \int_{\epsilon}^{\infty} dx \, x^{-3/2} \frac{1}{\sqrt{1+ a b \frac{\tan{\sqrt{a x}}}{\sqrt{a x}}}} \\+ i \epsilon^{-1/2} \int_{\pi}^{-\pi} d\phi \, e^{-i \phi/2} \frac{e^{\epsilon e^{i \phi} t}}{\sqrt{1+ a b \frac{\tanh{\sqrt{a \epsilon e^{i \phi}}}}{\sqrt{a \epsilon e^{i \phi}}}}}$$
Note that the first integral is finite as $\epsilon \to 0$. In this limit, the second integral approaches
$$- i 2 \epsilon^{-1/2} \int_{1}^{\infty} dy \, y^{-3/2} = -i 4 \epsilon^{-1/2}$$
The third integral approaches
$$i \epsilon^{-1/2} \int_{-\pi}^{\pi} d\phi \, e^{-i \phi/2} = +i 4 \epsilon^{-1/2}$$
Obviously, these cancel. Higher order expansions in $\epsilon$ vanish as $\epsilon \to 0$, so the ILT may now be written as
This is as far as I can take it. The integral on the RHS is well-defined for all values of $t$ but I imagine the presence of the $\tan$ function may give a numerical integrator some fits. Good luck!
ADDENDUM
The root in the denominator may also give rise to other branch points in the complex plane. Calculating these branch point locations seems to me to be quite difficult and maybe it is just as well to assume that $a$ and $b$ are defined such that there are no such branch points in the left half-plane (if that is even possible).