In what follows we assume $\Re(a)>0$ and $\Re(b)>0$. 

Begin with the case $a+b=k\in\mathbb N$. Using Pochhammer contour $P$, one can relate what's going on on $[0,1]$ to what is going on on a circle $C:=x_*\mathbb S^1$, $|x_*|>1$. Indeed, looking carefully at determinations of $f(z):=z^{a-1}(1-z)^{b-1}$ one has $$ \oint_Pf(z)dz = (1-\exp 2ib\pi)\oint_C f(z)dz ~~~(*)\\=(1-\exp 2ib\pi)(1-\exp 2ia\pi)\int_0^1f(z)dz$$ the last equality being given by Pochhammer formula, so that $$ \oint_C f(z)dz = 2i(-1)^{1-a}\sin(a\pi) \int_0^1f(z)dz .$$ Since $f(z)=z^{a+b-2}(1/z-1)^{b-1}$ is holomorphic near $\infty$ we have $$\oint_Cf(z)dz = -\oint_C f(1/x)\frac{dx}{x^2}=(-1)^{b-1}\oint_C x^{-k}(1-x)^{b-1}dx , $$ which is a contour integral. It can be evaluated by looking a the expansion of $$(1-x)^{b-1} = \sum_n \frac{\Gamma(b)}{\Gamma(n+1)\Gamma(b-n)}x^n .$$ The residue of $x^{-k}(1-x)^{b-1}$ at $0$ is obtained for $n+1=k$, that is $\frac{\Gamma(b)}{\Gamma(a+b)\Gamma(1-a)}$ which allows to conclude using Gamma reflection formula $\Gamma(1-a)\Gamma(a)\sin(a\pi)=\pi$.


The next step is to deal with the case $a+b=p/q\in\mathbb Q$, then conclude by analyticity and accumulation. This case is dealt with by taking a linear combination of $\oint_Cf(z)dz$ with weights $\exp (2in\pi/q)$ to obtain the same kind of relation as $(*)$. I'll write details later, but they should be straightforward.