why is the z transform transfer function 1/(z-1) called an integrator?
There are a couple reasons.  One is that $(1-z^{-1})$ represents $x[n]-x[n-1]$ which is a finite difference over a very small period of time.  and that is an approximation to a differentiator.  The reciprocal is 

$$ \frac{1}{1-z^{-1}} = \frac{z}{z-1} $$

which is the inverse operator.  We normally call the inverse operation of differentiation, we call that "integration".

Another reason is simply to implement that term as a transfer function of a tiny little LTI system:

$$ \frac{Y(z)}{X(z)} = \frac{1}{z-1} = \frac{z^{-1}}{1-z^{-1}} $$

or

$$ Y(z)(1 - z^{-1}) = Y(z) - Y(z) z^{-1} = X(z) z^{-1} $$

that translates to

$$ y[n] - y[n-1] = x[n-1] $$

or

$$ y[n] = y[n-1] + x[n-1] $$

so the current output sample is the previous output added to the (slightly delayed) input.  the output is an accumulation of the input.  similarly an integral of an input $x(t)$ is an accumulation of that input until the present time $t$.

i would say a better representation of a discrete-time integrator is $\frac{z}{z-1}$.  that would correspond to

$$ y[n] = y[n-1] + x[n] $$