If there wasn't the $x(t)$ term, I could use $x(t) = x$ and $x(t) = -x$ to disprove invertibility, but I can't think of two functions that give the same $y(t)$ in this case. When I tried proving invertibility using general $x_1(t)$ and $x_2(t)$, I get:
$$\left(x_1(t-1)\right)^2 - \left(x_2(t-1)\right)^2 + x_1(t) - x_2(t) + \left(x_1(t+1)\right)^2 - \left(x_2(t+1)\right)^2 =0 $$which leads me nowhere.
Let $x_1(t)=t-1/4,\ x_2(t)=-t-1/4.$ Both choices give $y(t)=2t^2+15/8.$
To get this I assumed $x(t)=at+b$ and plugged it in, realized the absolute value of $a$ was determined, then put $+1,-1$ for $a$ and different letters for $b$, and found one only needed $b_1+b_2=-1/2.$