Using complex integration over the direct unit circle $S^1$, one gets $$\mathbb P(X\gt k)=\frac1{2\mathrm i\pi}\oint_{S^1}\frac{1-G(z)}{1-z}\cdot\frac{\mathrm dz}{z^{k+1}}. $$
To prove this, one can rely on two ingredients:
(1) A distribution transform
For every nonnegative integer valued random variable $X$ integrable and not almost surely zero, there exists a random variable $Y$ such that, for every $k\geqslant0$, $$ \mathbb P(Y=k)=\frac{\mathbb P(X\gt k)}{\mathbb E(X)}. $$ This is because each RHS is nonnegative and their sum is $1$ thanks to the well-known formula $$ \sum_{k\geqslant0}\mathbb P(X\gt k)=\mathbb E(X). $$ If the PGF of $X$ is the function $G$ defined by $G(z)=\mathbb E(z^X)$, then $\mathbb E(X)=G'(1)$ and the PGF of $Y$ is the function $F$ defined by $$ F(z)=\mathbb E(z^Y)=\frac{1-G(z)}{\mathbb E(X)\cdot(1-z)}. $$ (2) The "extraction" of coefficients of an entire series through Cauchy formula
For every nonnegative integer valued random variable $Y$ with PGF $F$ and every $k\geqslant0$, $$ \mathbb P(Y=k)=\frac1{2\mathrm i\pi}\oint_{S^1}F(z)\cdot\frac{\mathrm dz}{z^{k+1}}. $$ To see this, expand $F$ as $F(z)=\sum\limits_{n\geqslant0}\mathbb P(Y=n)z^n$ and use the fact that, for every $n\geqslant0$, $$ \oint_{S^1}z^n\cdot\frac{\mathrm dz}{z^{k+1}}=\left\{\begin{array}{ccc}2\mathrm i\pi & \text{if} & n=k,\\ 0 & \text{if} & n\ne k.\end{array}\right. $$ Using (1) and (2) together, one gets $$ \frac{\mathbb P(X\gt k)}{\mathbb E(X)}=\frac1{2\mathrm i\pi}\oint_{S^1}\frac{1-G(z)}{\mathbb E(X)\cdot(1-z)}\cdot\frac{\mathrm dz}{z^{k+1}}, $$ which is equivalent to the desired formula.