Basic fact: For every positive $u$ and $v$, $\displaystyle\int\limits_0^1x^{u-1}(1-x)^{v-1}\,\mathrm dx=\frac{\Gamma(u)\Gamma(v)}{\Gamma(u+v)}$.
Thus the PDF of $\Pi$ is not what you wrote but a multiple of this, and, for every $y$ in $\{0,1\}$, $ L_{a,b}^y=\int_0^1f_{Y\mid\Pi}(y\mid x)\cdot f_\Pi(x)\,\mathrm dx=\int_0^1x^y(1-x)^{1-y}\cdot\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}x^{a-1}(1-x)^{b-1}\,\mathrm dx, $ that is, $ L_{a,b}^y=\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\cdot\frac{\Gamma(a+y)\Gamma(b+1-y)}{\Gamma(a+b+1)}. $ In other words, $L_{a,b}^0=\dfrac{b}{a+b}$ and $L_{a,b}^1=\dfrac{a}{a+b}$.