Making a cube root function analytic on $\mathbb{C}\backslash [1,3]$

Given that $$\frac{((z-1)(z-2)(z-3))'}{(z-1)(z-2)(z-3)}=\frac1{z-1}+\frac1{z-2}+\frac1{z-3}$$ you see that all the three residues of the integrand are equal to one, so their sum is $3$. The "upper" and "lower" integrals thus differ by $6\pi i$.

After multiplication by $1/3$ they differ by $2\pi i$, and after $\exp$ they agree. IOW $f(z)$ won't be continuous, but $\exp(f(z)/3)$ still is.

If you believe in the Monodromy theorem and are comfortable with the Riemann sphere $\overline{\mathbb C}$, the following short argument suffices: $$\Big((z-1)(z-2)(z-3)\Big)^{1/3} = z\Big((1-1/z)(1-2/z)(1-3/z) \Big)^{1/3}$$ Here $\Big((1-1/z)(1-2/z)(1-3/z) \Big)^{1/3}$ admits analytic continuation in $\overline{\mathbb C}\setminus [1,3]$, which is a simply-connected domain. Therefore, it has a single-valued branch there.

If you are not comfortable with the Riemann sphere, but still believe in the Monodromy theorem, then move the segment to infinity, for example by letting $w=1/(z-1)$. This transforms $\overline{\mathbb C}\setminus [1,3]$ into $ {\mathbb C}\setminus [1/2,\infty)$. We get $$\Big((z-1)(z-2)(z-3)\Big)^{1/3} = \Big( w^{-1}(w^{-1}-1) (w^{-1}-2) \Big)^{1/3} =w^{-1} ( (1-w)(1-2w) )^{1/3} $$ Here $( (1-w)(1-2w) )^{1/3}$ admits analytic continuation in $ {\mathbb C}\setminus [1/2,\infty)$, which is a simply-connected domain. Therefore, it has a single-valued branch there.