For $0<a<1\land0<b$,
$$\small{\mathcal{I}{\left(a,b\right)}=-\pi\arcsin{\left(\sqrt{\frac{a}{a+b}}\right)}+\frac{\pi a}{\sqrt{a+b}}F_{3}{\left(\frac12,\frac12;\frac12,\frac12;2;a,\frac{a}{a+b}\right)}.}$$
Proof:
The inverse sine function may be defined on the complex unit circle via the integral representation
$$\arcsin{\left(z\right)}:=\int_{0}^{z}\frac{\mathrm{d}t}{\sqrt{1-t^{2}}};~~~\small{z\in\mathbb{C}\land\left|z\right|\le1}.$$
Using the integral representation of $\arcsin{\left(z\right)}$ above, we may express $\mathcal{I}{\left(a,b\right)}$ equivalently as a double integral with an algebraic integrand. It has come to my attention that the resulting double integral has a natural expression an Appell hypergeometric function of the third kind:
$$F_{3}{\left(\alpha,\alpha^{\prime};\beta,\beta^{\prime};\gamma;z,w\right)}=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\alpha^{\prime}\right)_{k}\left(\beta\right)_{n}\left(\beta^{\prime}\right)_{k}}{\left(\gamma\right)_{n+k}n!\,k!}z^{n}w^{k},$$
where $\left|z\right|<1\land\left|w\right|<1$.
An integral representation for $F_{3}$ is derived in the appendix below.
Now, assuming $0<a<1\land0<b$, we find
$$\begin{align}
\mathcal{I}{\left(a,b\right)}
&=\int_{0}^{a}\frac{\arcsin{\left(2t-1\right)}}{\sqrt{\left(a-t\right)\left(b+t\right)}}\,\mathrm{d}t\\
&=\int_{0}^{a}\mathrm{d}t\,\frac{1}{\sqrt{\left(a-t\right)\left(b+t\right)}}\int_{\frac12}^{t}\mathrm{d}u\,\frac{1}{\sqrt{u\left(1-u\right)}}\\
&=-\int_{0}^{a}\mathrm{d}t\,\frac{1}{\sqrt{\left(a-t\right)\left(b+t\right)}}\int_{0}^{\frac12}\mathrm{d}u\,\frac{1}{\sqrt{u\left(1-u\right)}}\\
&~~~~~+\int_{0}^{a}\mathrm{d}t\,\frac{1}{\sqrt{\left(a-t\right)\left(b+t\right)}}\int_{0}^{t}\mathrm{d}u\,\frac{1}{\sqrt{u\left(1-u\right)}}\\
&=-\frac{\pi}{2}\int_{0}^{a}\frac{\mathrm{d}t}{\sqrt{\left(a-t\right)\left(b+t\right)}}\\
&~~~~~+\int_{0}^{a}\mathrm{d}t\,\frac{1}{\sqrt{\left(a-t\right)\left(b+t\right)}}\int_{0}^{t}\mathrm{d}u\,\frac{1}{\sqrt{u\left(1-u\right)}}\\
&=-\frac{\pi}{2}\int_{0}^{a}\frac{\mathrm{d}u}{\sqrt{u\left(a+b-u\right)}};~~~\small{\left[a-t=u\right]}\\
&~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{a}{\sqrt{ax\left(a+b-ax\right)}}\int_{0}^{a-ax}\mathrm{d}u\,\frac{1}{\sqrt{u\left(1-u\right)}};~~~\small{\left[\frac{a-t}{a}=x\right]}\\
&=-\frac{\pi}{2}\int_{0}^{\frac{a}{a+b}}\frac{\mathrm{d}v}{\sqrt{v\left(1-v\right)}};~~~\small{\left[u=\left(a+b\right)v\right]}\\
&~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{a}{\sqrt{ax\left(a+b-ax\right)}}\int_{0}^{1-x}\mathrm{d}y\,\frac{a}{\sqrt{ay\left(1-ay\right)}};~~~\small{\left[u=ay\right]}\\
&=-\pi\int_{0}^{\sqrt{\frac{a}{a+b}}}\frac{\mathrm{d}w}{\sqrt{1-w^2}};~~~\small{\left[\sqrt{v}=w\right]}\\
&~~~~~+\frac{a}{\sqrt{a+b}}\int_{0}^{1}\mathrm{d}x\int_{0}^{1-x}\mathrm{d}y\,\frac{1}{\sqrt{yx\left(1-ay\right)\left(1-\frac{a}{a+b}x\right)}}\\
&=-\pi\arcsin{\left(\sqrt{\frac{a}{a+b}}\right)}+\frac{\pi a}{\sqrt{a+b}}F_{3}{\left(\frac12,\frac12;\frac12,\frac12;2;a,\frac{a}{a+b}\right)}.\\
\end{align}$$
Appendix:
Given the parameter assumptions
$$\small{z,w\in\mathbb{D}\land\alpha,\alpha^{\prime},\beta,\beta^{\prime},\gamma\in\mathbb{C}\setminus\mathbb{Z}^{\le0}\land0<\Re{\left(\beta\right)}\land0<\Re{\left(\beta^{\prime}\right)}\land0<\Re{\left(\gamma-\beta-\beta^{\prime}\right)}}$$
an integral representation for the Appell hypergeometric function of the third kind may be derived as follows:
$$\begin{align}
F_{3}
&=F_{3}{\left(\alpha,\alpha^{\prime};\beta,\beta^{\prime};\gamma;z,w\right)}\\
&=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\alpha^{\prime}\right)_{k}\left(\beta\right)_{n}\left(\beta^{\prime}\right)_{k}}{\left(\gamma\right)_{n+k}n!\,k!}z^{n}w^{k}\\
&=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\alpha^{\prime}\right)_{k}\left(\beta\right)_{n}\left(\beta^{\prime}\right)_{k}}{\left(\gamma\right)_{n}\left(\gamma+n\right)_{k}n!\,k!}z^{n}w^{k}\\
&=\sum_{n=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\beta\right)_{n}z^{n}}{\left(\gamma\right)_{n}n!}\sum_{k=0}^{\infty}\frac{\left(\alpha^{\prime}\right)_{k}\left(\beta^{\prime}\right)_{k}w^{k}}{\left(\gamma+n\right)_{k}\,k!}\\
&=\sum_{n=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\beta\right)_{n}z^{n}}{\left(\gamma\right)_{n}n!}\,{_2F_1}{\left(\alpha^{\prime},\beta^{\prime};\gamma+n;w\right)}\\
&=\small{\sum_{n=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\beta\right)_{n}z^{n}}{\left(\gamma\right)_{n}n!}\cdot\frac{1}{\operatorname{B}{\left(\beta^{\prime},\gamma+n-\beta^{\prime}\right)}}\int_{0}^{1}\frac{t^{\beta^{\prime}-1}\left(1-t\right)^{\gamma+n-\beta^{\prime}-1}}{\left(1-wt\right)^{\alpha^{\prime}}}\,\mathrm{d}t}\\
&=\small{\sum_{n=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\beta\right)_{n}z^{n}}{\left(\gamma\right)_{n}n!}\cdot\frac{\Gamma{\left(\gamma+n\right)}}{\Gamma{\left(\beta^{\prime}\right)}\,\Gamma{\left(\gamma+n-\beta^{\prime}\right)}}\int_{0}^{1}\frac{t^{\beta^{\prime}-1}\left(1-t\right)^{\gamma+n-\beta^{\prime}-1}}{\left(1-wt\right)^{\alpha^{\prime}}}\,\mathrm{d}t}\\
&=\small{\frac{\Gamma{\left(\gamma\right)}}{\Gamma{\left(\beta^{\prime}\right)}\,\Gamma{\left(\gamma-\beta^{\prime}\right)}}\sum_{n=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\beta\right)_{n}z^{n}}{\left(\gamma-\beta^{\prime}\right)_{n}n!}\int_{0}^{1}\frac{t^{\beta^{\prime}-1}\left(1-t\right)^{\gamma+n-\beta^{\prime}-1}}{\left(1-wt\right)^{\alpha^{\prime}}}\,\mathrm{d}t}\\
&=\small{\frac{1}{\operatorname{B}{\left(\beta^{\prime},\gamma-\beta^{\prime}\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{t^{\beta^{\prime}-1}\left(1-t\right)^{\gamma-\beta^{\prime}-1}}{\left(1-wt\right)^{\alpha^{\prime}}}\sum_{n=0}^{\infty}\frac{\left(\alpha\right)_{n}\left(\beta\right)_{n}z^{n}\left(1-t\right)^{n}}{\left(\gamma-\beta^{\prime}\right)_{n}n!}}\\
&=\small{\frac{1}{\operatorname{B}{\left(\beta^{\prime},\gamma-\beta^{\prime}\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{t^{\beta^{\prime}-1}\left(1-t\right)^{\gamma-\beta^{\prime}-1}}{\left(1-wt\right)^{\alpha^{\prime}}}{_2F_1}{\left(\alpha,\beta;\gamma-\beta^{\prime};z\left(1-t\right)\right)}}\\
&=\frac{1}{\operatorname{B}{\left(\beta^{\prime},\gamma-\beta^{\prime}\right)}\cdot\operatorname{B}{\left(\beta,\gamma-\beta-\beta^{\prime}\right)}}\times\\
&~~~~~\int_{0}^{1}\mathrm{d}t\,\frac{t^{\beta^{\prime}-1}\left(1-t\right)^{\gamma-\beta^{\prime}-1}}{\left(1-wt\right)^{\alpha^{\prime}}}\int_{0}^{1}\mathrm{d}u\,\frac{u^{\beta-1}\left(1-u\right)^{\gamma-\beta-\beta^{\prime}-1}}{\left(1-z\left(1-t\right)u\right)^{\alpha}}\\
&=\frac{\Gamma{\left(\gamma\right)}}{\Gamma{\left(\beta\right)}\,\Gamma{\left(\beta^{\prime}\right)}\,\Gamma{\left(\gamma-\beta-\beta^{\prime}\right)}}\times\\
&~~~~~\small{\int_{0}^{1}\mathrm{d}t\,\frac{t^{\beta^{\prime}-1}}{\left(1-wt\right)^{\alpha^{\prime}}}\int_{0}^{1-t}\mathrm{d}y\,\frac{y^{\beta-1}\left(1-t-y\right)^{\gamma-\beta-\beta^{\prime}-1}}{\left(1-zy\right)^{\alpha}}};~~~\small{\left[\left(1-t\right)u=y\right]}\\
&=\frac{\Gamma{\left(\gamma\right)}}{\Gamma{\left(\beta\right)}\,\Gamma{\left(\beta^{\prime}\right)}\,\Gamma{\left(\gamma-\beta-\beta^{\prime}\right)}}\times\\
&~~~~~\int_{0}^{1}\mathrm{d}x\int_{0}^{1-x}\mathrm{d}y\,\frac{x^{\beta^{\prime}-1}y^{\beta-1}\left(1-x-y\right)^{\gamma-\beta-\beta^{\prime}-1}}{\left(1-wx\right)^{\alpha^{\prime}}\left(1-zy\right)^{\alpha}}.\blacksquare\\
\end{align}$$