I assume $b\notin\mathbb{Z}_{\leqslant 0}$ everywhere, and $_1F_1(a,b,z)$ defined by $$_1F_1(a,b,z)=1+\frac{a}{b}\frac{z}{1!}+\frac{a(a+1)}{b(b+1)}\frac{z^2}{2!}+\frac{a(a+1)(a+2)}{b(b+1)(b+2)}\frac{z^3}{3!}+\ldots$$
$\color{red}{(1)}$ is claimed to hold if $b$ is a positive integer (this is stated a few lines above the equation). We write $$I_1:=\frac{\Gamma(b)}{2\pi i}\oint_{C_1}e^t t^{a-b}(t-z)^{-a}dt=\frac{(b-1)!}{2\pi i}\oint_{C_1}\frac{e^t}{t^b}\left(1-\frac{z}{t}\right)^{-a}dt$$ (now we see that the notion of "branch cut from $t=0$ to $t=z$" makes sense), deform $C_1$ into a large circle centered at $0$ (to have $\sup_t|z/t|<1$ on it), and use the binomial series for $(1-z/t)^{-a}$ (which then converges uniformly on $C_1$): $$I_1=\frac{(b-1)!}{2\pi i}\sum_{n=0}^\infty\frac{a\cdots(a+n-1)}{n!}z^n\oint_{C_1}\frac{e^t\,dt}{t^{b+n}}=\sum_{n=0}^\infty\frac{a\cdots(a+n-1)}{b\cdots(b+n-1)}\frac{z^n}{n!}$$ (the last integral is equal to $2\pi i/(b+n-1)!$ by Cauchy's theorem).
$\color{red}{(2)}$ is claimed to hold if $b$ is a positive integer and if $a$ is not an integer. The contour $C_2$ is supposed to encircle the branch cut of the integrand (again, joining $t=0$ and $t=1$; the segment $[0,1]$ of the real axis can be taken for that). This time, we can prove it for $0<\Re a<b$ (with $a$ still not an integer) and rely on analytic continuation elsewhere. To do so, we squeeze $C_2$ so that it encircles $[0,1]$ closely, and assume the integrand takes real values on the lower edge of the cut (this is suggested by the way the contour is depicted in the book); then, on the upper edge, it gains the factor of $e^{-2a\pi i}$, and in the limit of "closely", we obtain $$\oint_{C_2}e^{zt}t^{a-1}(1-t)^{b-a-1}dt=(1-e^{-2a\pi i})\int_0^1 e^{zt}t^{a-1}(1-t)^{b-a-1}dt.$$ The (well-known) last integral is evaluated using $e^{zt}=\sum_{n=0}^\infty(zt)^n/n!$ and equals $$\sum_{n=0}^\infty\frac{z^n}{n!}\frac{\Gamma(a+n)\Gamma(b-a)}{\Gamma(b+n)}=\frac{\Gamma(a)\Gamma(b-a)}{\Gamma(b)}{}_1F_1(a,b,z).$$
Finally, the corrected version of $\color{red}{(3)}$ (see below) holds for $\Re a>0$ and $b-a$ not a positive integer. Use $$\int_0^{(1+)}t^{\alpha-1}(t-1)^{\beta-1}dt=2i\sin\beta\pi\ \mathrm{B}(\alpha,\beta)\qquad(\Re\alpha>0)$$ stated here; the contour starts at $t=0$, encircles $t=1$ counterclockwise, and returns to $t=0$, thus crossing the branch cut of $(t-1)^{\beta-1}$ at this point. This formula is proven much like we did the $(2)$ above. Now we get the same way $$\int_0^{(1+)}e^{zt}t^{a-1}(t-1)^{b-a-1}dt=2i\sin(b-a)\pi\ \mathrm{B}(a,b-a)\ _1F_1(a,b,z),$$ or, using the reflection formula $\Gamma(b-a)\sin(b-a)\pi=\pi/\Gamma(1+a-b)$, $$_1F_1(a,b,z)=\frac{\Gamma(b)\Gamma(1+a-b)}{2\pi i\Gamma(a)}\int_0^{(1+)}e^{zt}t^{a-1}(t-1)^{b-a-1}dt.$$
