Equality in distribution based on particular equalities in expected values or in series of moments

Consider the two parameter family of distributions which place mass $1/4$ at the values $1/2 pm alpha$ and $1/2 pm beta$ for $1/2 ge beta ge alpha ge 0$.

For such a family, the means are all $1/2,$ while $$ mathbb{E}_{alpha, beta} [(1-X) ln (1-X)] = -frac{log 2}{2} + frac{(1-2alpha) log (1-2alpha) + (1+ 2alpha) log(1+2alpha)}{8} \ + frac{(1-2beta) log (1-2beta) + (1+ 2beta) log(1+2beta)}{8} \ =: -log2/2 + f(alpha, beta)/8.$$

If we can show that there are two distinct $(alpha, beta)$ for which $f$ takes the same value, then we would demonstrate a counterexample - $X$ and $Y$ can be distributed according to these two different laws.

Now $f(alpha, beta) = g(alpha) + g(beta)$ where $$g(x) = (1-2x) log(1-2x) + (1+2x) log (1+2x).$$ Notice that over the domain $[0,1/2],$ $g$ is a continuous monotonically increasing functinon, with $g(0) = 0$ and $g(1/2) = 2log 2$. This suggests that for any pair $x < y$ and small enough $delta > 0$, we should be able to find $varepsilon$ such that $$ g(x) + g(y) = g(x + delta) + g(y-varepsilon).$$

We, of course, need much less than this claim - we only need this for one value of $(alpha, beta)$. I'll study $(0,1/2)$. Note that $f(0, 1/2) = 2log 2$. Pick $0 < delta < 1/2$ to be a number such that $0 < g(delta) < log 2$ (this exists due to the intermediate value theorem). I claim that there must exist a $varepsilon < 1/2 - delta$ such that $ f(delta, 1/2 - varepsilon) = g(1/2 - varepsilon) + g(delta) = 2log(2).$

Indeed, consider the function $$ h(varepsilon) = f(delta, 1/2 - varepsilon) -2log 2 = g(1/2 - varepsilon) + g(delta) - 2log 2.$$ Then $$ h(0) = g(1/2) + g(delta) - 2log 2 > 0$$ and $$ h(1/2 - delta) = 2g(delta) - 2log 2 < 0.$$ The claim follows by the continuity of $h$ and the intermediate value theorem.