It would be the situation in B: you would deform around the pole. It works as follows.
The inverse Laplace transform is given by Cauchy's theorem. I present the parametrization of each piece of the contour, assuming that the radius of the semicircular detour about the pole $z=-1$ and the branch point $z=0$ is $\epsilon$:
$$\int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} + e^{i \pi} \int_{\infty}^{1+\epsilon} dx \frac{e^{-t x}}{e^{i \pi/2} \sqrt{x} (1-x)}+i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{t (-1+ \epsilon e^{i \phi})}}{\sqrt{e^{i \pi}+\epsilon e^{i \phi}} (\epsilon e^{i \phi})}\\+ e^{i \pi} \int_{1-\epsilon}^{\epsilon} dx \frac{e^{-t x}}{e^{i \pi/2} \sqrt{x} (1-x)}+i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac{e^{t \epsilon e^{i \phi}}}{\sqrt{\epsilon e^{i \phi}} (1+\epsilon e^{i \phi})} +e^{-i \pi} \int_{\epsilon}^{1-\epsilon} dx \frac{e^{-t x}}{e^{-i \pi/2} \sqrt{x} (1-x)}\\+ i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{e^{t (-1+ \epsilon e^{i \phi})}}{\sqrt{e^{-i \pi}+\epsilon e^{i \phi}} (\epsilon e^{i \phi})}+ e^{-i \pi} \int_{1+\epsilon}^{\infty} dx \frac{e^{-t x}}{e^{-i \pi/2} \sqrt{x} (1-x)} = 0$$
Note that the integrals about the semicircular detours above and below the axis (the 3rd and the 7th integrals) cancel. In the limit as $\epsilon \to 0$, the integral about the branch point (the 5th integral) also vanishes. We are then left with, as $\epsilon \to 0$,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} + \frac1{2 \pi} PV \int_{\infty}^0 dx \frac{e^{-t x}}{\sqrt{x} (1-x)} - \frac1{2 \pi} PV \int_0^{\infty} dx \frac{e^{-t x}}{\sqrt{x} (1-x)} = 0$$
where $PV$ denotes the Cauchy principal value of the integral. Thus, the ILT is given by (subbing $x=u^2$)
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = \frac1{\pi} PV \int_{-\infty}^{\infty} du \, \frac{e^{-t u^2}}{1-u^2} $$
To evaluate the integral, we rewrite as
$$e^{-t} PV \int_{-\infty}^{\infty} du \, \frac{e^{t (1- u^2)}}{1-u^2} = e^{-t} I(t)$$
where
$$I'(t) = e^{t} PV \int_{-\infty}^{\infty} du \, e^{-t u^2} = \sqrt{\pi} t^{-1/2} e^{t} $$
and $I(0) = 0$. Thus,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = e^{-t}\frac1{\pi} \sqrt{\pi} \int_0^t dt' \, t'^{-1/2} e^{t'} = e^{-t} \frac{2}{\sqrt{\pi}} \int_0^{\sqrt{t}} dv \, e^{v^2} $$
or, finally,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = e^{-t} \operatorname{erfi}{\left (\sqrt{t} \right )} $$
My choice of the branch cut on the negative real axis means the jump in phase is between $\pi$ and $-\pi$. Here, $1+z^2$ has a phase of $\pi$ on the right edge of the branch cut and a phase of $-\pi$ on the left side of the branch cut. Correspondingly,
$$0 = I + PV \int_{\sqrt{a}}^\infty i dy \frac{2\pi i}{1-y^2} + (-\pi \ln(a-1)/2 \color{red}{+ \pi i (-\pi i)/(2i)}) + (-\pi \ln(a-1)/2 \color{red}{- \pi i (-\pi i)/(2i)})$$
This resolves the discrepancy, as the two contributions in red cancel.
The reason I made my mistake is that the difference across a branch cut of a log is $2 \pi i$, and I'm quite used to taking my branch cuts on the positive real axis for which the phase goes from $0$ to $2\pi$, so I stuck in a $2\pi i$ without thinking.
This more careful bookkeeping unearthed a second mistake I made in the question. The contribution $\int_{\sqrt{a}}^\infty i dy \frac{2\pi i}{1-y^2}$ should instead be $\int_{\sqrt{a}}^\infty i dy \frac{-2\pi i}{1-y^2}$, since the left hand side of the branch cut has a phase that is lower by $2 \pi i$ relative to the right hand side, not higher.
Everything works out in the end, since $\int_{\sqrt{a}}^\infty i dy \frac{2\pi i}{1-y^2} = -\pi \ln(\frac{\sqrt{a}-1}{\sqrt{a}+1})$, which is negative what I had written; two wrongs had made a right in my first calculation.
Best Answer
Let us state our claim:
$$\int_{-1}^1 \frac{1}{a+bz} \frac{1}{\sqrt{1-z^2}} \; dz = \frac{\pi}{\sqrt{a^2-b^2}}.$$
To start we introduce a dogbone contour as our contour with two circles at either end parameterized by $-1 + \varepsilon \exp(i\theta)$ where $\psi \le \theta \le 2\pi-\psi$ and $1 + \varepsilon \exp(i\theta)$ where $-\pi+\psi\le \theta \le \pi-\psi$ Here $\psi$ is negligeably small and the ends of the arcs are not closed so as to leave a small opening. We connect the two terminal vertices above the real axis by a line, the same for the two vertices below. These will produce multiples of the desired integral in the limit.
We will use
$$f(z) = \frac{1}{a+bz} \exp(-1/2 \times\mathrm{LogA}(1+z)) \exp(-1/2 \times\mathrm{LogB}(1-z)).$$
Here $\mathrm{LogA}$ denotes the branch of the logarithm where $0 \le \arg \mathrm{LogA} \lt 2\pi$ and $\mathrm{LogB}$ where $-\pi \le \arg \mathrm{LogB} \lt \pi.$ The branch cut from $\mathrm{LogA}$ includes the interval $[-1,1]$ on the real axis which is inside the dogbone contour while for $x\gt 1$ both branch cuts apply. In fact in the latter case they cancel and we have continuity across $(1,\infty)$ and may derive analyticity by Morera's theorem as explained at this MSE link. The pole at $z = -a/b$ is not on either cut with $a\gt b \gt 0.$
For the continuity the rational factor is obviously the same above and below the cut, while for the two logarithmic factors we get for $x\gt 1$ and above the cut $\mathrm{LogA}(1+x) = \log(x+1)$ and $\mathrm{LogB}(1-x) = \log(x-1)-\pi i$ (rotation) and below the cut $\mathrm{LogA}(1+x) = \log(x+1) + 2\pi i$ and $\mathrm{LogB}(1-x) = \log(x-1) + \pi i$ (rotation again). This yields above the cut
$$\exp(-1/2\times \log(x+1))\exp(-1/2\times(\log(x-1)-\pi i)) \\ = \frac{1}{\sqrt{x^2-1}} \exp(1/2 \times \pi i) = \frac{i}{\sqrt{x^2-1}}$$
and below the cut
$$\exp(-1/2\times(\log(x+1)+2\pi i))\exp(-1/2\times(\log(x-1)+\pi i)) \\ = \frac{1}{\sqrt{x^2-1}} \exp(-1/2 \times 3\pi i) = \frac{i}{\sqrt{x^2-1}}$$
and we have continuity across the cut. (For the cut itself the values are from the above-the-cut case.) By Morera we have analyticity in $\mathbb{C}\backslash [-1, 1]$ for the square root term. The value may be simplified to $1/\sqrt{1-x^2}$ if desired. Next we need to verify that the two segments above and below the single branch cut enclosed by the dogbone contour (i.e., the line connecting $-1$ to $1$) are multiples of the target integral. We get above the cut
$$\exp(-1/2\times(\log(x+1))\exp(-1/2\times(\log(1-x)) = \frac{1}{\sqrt{1-x^2}}$$
and below
$$\exp(-1/2\times(\log(x+1)+2\pi i))\exp(-1/2\times(\log(1-x))) = -\frac{1}{\sqrt{1-x^2}}.$$
This means that with a counter-clockwise traversal of the contour we pick up the target integral times a factor of $-2.$ We will compute the integral using the poles outside of the dogbone contour. Next to compute the residues we have that the residue at infinity is zero by inspection. We get for the pole at $-a/b$ that the residue is
$$\;\underset{z=-a/b}{\mathrm{res}}\; f(z) = \;\underset{z=-a/b}{\mathrm{res}}\; \frac{1}{b} \frac{1}{z+a/b} \frac{1}{\sqrt{1-z^2}} \\ = \frac{1}{b} \frac{1}{\sqrt{1-a^2/b^2}} = \frac{1}{\sqrt{b^2-a^2}}.$$
Now the contour produces twice the desired value as explained earlier and hence it is given by (poles outside rather than inside contour)
$$- \frac{1}{2} \times - 2\pi i \times \frac{1}{\sqrt{b^2-a^2}} = \frac{\pi i}{\sqrt{b^2-a^2}}$$
which is
$$\bbox[5px,border:2px solid #00A000]{ \frac{\pi}{\sqrt{a^2-b^2}}.}$$
To be rigorous we also need to show that the contribution from the circular arcs vanishes in the limit. We get for the left arc with $z = -1 + \varepsilon \exp(i\theta)$ where $0\le\theta\lt 2\pi$ (the arc does not cross any cuts of the two logarithms):
$$\frac{1}{a-b+b\varepsilon\exp(i\theta)} \\ \times \exp(-1/2\times \mathrm{LogA}(\varepsilon \exp(i\theta))) \exp(-1/2\times \mathrm{LogB}(2-\varepsilon \exp(i\theta))) \\ = \frac{1}{a-b+b\varepsilon\exp(i\theta)} \\ \times \exp(-1/2\times (\log\varepsilon + i\theta)) \exp(-1/2\times \mathrm{LogB}(2-\varepsilon \exp(i\theta))).$$
The middle term is $\exp(- i\theta/2) \frac{1}{\sqrt{\varepsilon}}$ and the other two are constant in the asymptotic as $\varepsilon\rightarrow 0$. Multiply by $2\pi \varepsilon$ to have it vanish. We get for the right arc with $z = 1 + \varepsilon \exp(i\theta)$ where $-\pi\le\theta\lt\pi$ (this arc crosses both cuts, so we have to be careful):
$$\frac{1}{a+b+b\varepsilon\exp(i\theta)} \\ \times \exp(-1/2\times \mathrm{LogA}(2+\varepsilon \exp(i\theta))) \exp(-1/2\times \mathrm{LogB}(-\varepsilon \exp(i\theta))).$$
We may switch this to
$$\frac{1}{a+b+b\varepsilon\exp(i\theta)} \\ \times \exp(-1/2\times \Re \mathrm{LogA}(2+\varepsilon \exp(i\theta))) \exp(-1/2\times \Re \mathrm{LogB}(-\varepsilon \exp(i\theta))).$$
This is because the two imaginary components from the two logarithms produce modulus one, being situated on the unit circle. We obtain $\frac{1}{\sqrt{\varepsilon}}$ from the third term and the other two are once more constant in the asymptotic. Hence the contribution from this arc will vanish as well on multiplication by $2\pi\varepsilon.$ These two arcs were in fact indented by $\psi$ which we have going to zero.
This concludes the argument.