I guess the source of the question would be this, which contains a similar problem:
Let $U_1$ and $U_2$ be two independent random variables, each with uniform distribution on $[0,1]$. Let the random variable $β_{a, b}$ have $\text{Beta}(a,b)$ distribution, and let $B_p$ be a $\text{Bernoulli}(p)$ random variable that is independent of $β_{a,b}$. Find $p$, $a$, and $b$ so that $U_2−U_1$ has the same distribution as $(2B_p−1)β_{a, b}$.
Work/Question: By noticing $U_2-U_1$ is the convolution of a uniform on $[0,1]$ and another on $[-1,0]$, we can easily get the distribution as a triangular one i.e.
$$F_{U_2-U_1}(x)=\begin{cases}0 & x<-1 \\ \frac{x^2}{2}+x+\frac{1}{2} & x\in[-1,0) \\ -\frac{x^2}{2}+x+\frac{1}{2} & x\in[0,1)\end{cases}.$$
For the latter, $2B_p - 1$ is 1 with probability $p$ and $-1$ with probability $1-p$, so
$$F_{(2B_p−1)β_{a, b}}(x)=\mathbb{P}\left((2B_p−1)β_{a, b}\leq x,2B_p-1=1\right)+\mathbb{P}\left((2B_p−1)β_{a, b}\leq x, 2B_p - 1 = - 1\right)$$
$$=pF_{\beta_{a, b}}(x)+(1-p)\left(1 - F_{\beta_{a,b}}(-x)\right)$$
$$=\begin{cases} 0 & x<-1 \\ (1-p)\left(1 - F_{\beta_{a,b}}(-x)\right) & x\in[-1, 0) \\ (1-p) + pF_{\beta_{a, b}}(x) & x\in[0,1) \end{cases}.$$
Now equipped with the distributions, I figure we would just equate them and solve for the three. By inspection, I believe $p=\frac{1}{2}$, $a=1$, and $b=2$ solves our problem. To verify this, the latter distribution has for $x\in [-1,0)$
$$\frac{1}{2}\left(1 - \int\limits_0^{-x} \frac{\Gamma (3)}{\Gamma (1)\Gamma (2)} y^0(1-y)^{1}\;dy\right)=\frac{1}{2}\left(1-2(-x-\frac{x^2}{2})\right)=\frac{x^2}{2}+x+\frac{1}{2}$$
as desired and similarly the other equation holds. My only concern is that I feel like this system is underdetermined as there are two equations with three unknowns. Or does $p+(1-p)=1$ and $p\in[0,1]$ constitute a third equation from which this would be considered unique?
Your solution is the obvious one, and indeed the only one, as
$U_2−U_1$ is a triangular distribution on $[-1,1]$ symmetric about the mode of $0$.
Since $2B_p-1$ is $\pm1$, you are going to need $p=\frac12$ to make positive and negative outcomes equally likely.
$|U_2−U_1|$ has a triangular distribution on $[0,1]$ with a mode of $0$, which you need to match to the distribution of $\beta_{a, b}$ is a $\text{Beta}(1,2)$ distribution, i.e. $a=1, b=2$.
In terms of your analysis $p=\frac12$ is the only possibility for $F_{(2B_p−1)β_{a, b}}(0) = F_{U_2-U_1}(0) =\frac12.$
You can find $a$ and $b$ if you differentiate your CDF to give a density of $f_{|U_2-U_1|}=2-2x=2x^0(1-x)^1$ while for a Beta distribution $f_{\beta_{a, b}} \propto x^{a-1}(1-x)^{b-1}$ and matching coefficients gives your result uniquely.