I think I found an answer, but I would appreciate if someone would check to see if I did this correctly before I mark this question as answered.
Under Series formulas for the Incomplete Beta Function, Wolfram gives that
$$B(z; a, b) \propto \frac{z^a}{a}\left(1+O(z)\right)$$
and under Approximation for the Beta Function Wikipedia gives that
$$B(a, b) \propto \Gamma(b)a^{-b}$$
when $a$ grows and $b$ stays fixed (credit to Brevan Ellefsen for finding this second approximation). Because of these two approximations we can write our limit as
$$\lim_{n\to\infty} nx^{n-1}\frac{C\frac{\left(1-\frac{x^2}{4}\right)^{\frac{n+1}{2}}}{\frac{n+1}{2}}}{\Gamma\left(\frac{1}{2}\right)\left(\frac{n+1}{2}\right)^{-\frac{1}{2}}}$$
where $C$ is some constant not depending on $n$.
$$\lim_{n\to\infty} nx^{n-1}\frac{C\left(1-\frac{x^2}{4}\right)^{\frac{n+1}{2}}}{\sqrt{\pi}\sqrt{\frac{n+1}{2}}}$$
$$c\lim_{n\to\infty} \frac{n}{\sqrt{n+1}}x^{n-1}\left(1-\frac{x^2}{4}\right)^{\frac{n+1}{2}}$$
for some other constant $c$.
Let $x = 2\sin\theta$. The limit then becomes
$$c\lim_{n\to\infty} \frac{n}{\sqrt{n+1}}\left(2\sin\theta\right)^{n-1}\left(\cos\theta\right)^{n+1}$$
$$c\lim_{n\to\infty} \frac{n}{\sqrt{n+1}}\left(2\sin\theta\cos\theta\right)^{n-1}\cos^2\theta$$
$$c\lim_{n\to\infty} \frac{n}{\sqrt{n+1}}\left(\sin\left(2\theta\right)\right)^{n-1}\cos^2\theta$$
Note that, whenever $\left|\sin\left(2\theta\right)\right| < 1$, the $\left(\sin\left(2\theta\right)\right)^{n-1}$ term tends to $0$ more quickly than the $\frac{n}{\sqrt{n+1}}$ term tends to $\infty$ (since $\frac{n}{\sqrt{n+1}}$ is asymptotically a power of $n$ and $\left(\sin\left(2\theta\right)\right)^{n-1}$ is exponential), and the limit is just $0$; also note that when $\left|\sin\left(2\theta\right)\right| = 1$ the limit tends to $\infty$. Thus, this reduces to the problem of finding $\theta$ such that
$$\sin^2\left(2\theta\right) = 1$$
$$4\sin^2\theta\cos^2\theta = 1$$
$$x^2\left(1-\frac{x^2}{4}\right) = 1$$
$$4x^2-x^4 = 4$$
$$x^2 = 2$$
$$x = \pm\sqrt{2}$$
So, at these points the limit tends to $\infty$ whereas everywhere else it tends to $0$.
Best Answer
Using the definitions you wrote $$\text{lhs}=\frac{B_x(a,a)}{B(a,a)}$$ $$\text{rhs}=1-\frac{B_{4 (1-x) x}\left(a,\frac{1}{2}\right)}{2 B\left(a,\frac{1}{2}\right)}$$ Now, for $ \frac 12 \leq x \leq 1$ $$\frac d {dx}\text{[lhs-rhs]}=\frac{(1-x)^{a-1} x^{a-1}}{B(a,a)}-\frac{2^{2 a-3} (4 (1-x)-4 x) ((1-x) x)^{a-1}}{\sqrt{1-4 (1-x) x} B\left(a,\frac{1}{2}\right)}$$ Since $ \frac 12 \leq x \leq 1$, the last term simplify and write $$\frac{2^{2 a-1} ((1-x) x)^{a-1}}{B\left(a,\frac{1}{2}\right)}$$ $$\frac d {dx}\text{[lhs-rhs]}=\frac{(1-x)^{a-1} x^{a-1}}{B(a,a)}-\frac{2^{2 a-1} ((1-x) x)^{a-1}}{B\left(a,\frac{1}{2}\right)}$$ Divide by $(1-x)^{a-1} x^{a-1}$ to get $$\frac{\frac d {dx}\text{[lhs-rhs]} } {(1-x)^{a-1} x^{a-1} }=\frac{1}{B(a,a)}-\frac{2^{2 a-1}}{B\left(a,\frac{1}{2}\right)}=0$$ by the properties of the beta function.
So, for $ \frac 12 \leq x \leq 1$, $\text{[lhs-rhs]}$ is a constant. For $x=1$ for example $$\text{lhs}=\frac{B_1(a,a)}{B(a,a)}=1 \qquad \text{and} \qquad \text{rhs}=1-\frac{\Gamma \left(a+\frac{1}{2}\right) B_0\left(a,\frac{1}{2}\right)}{2 \sqrt{\pi } \Gamma (a)}=1$$ Then, the identity.