Multiply the fraction by $1$ in the carefully chosen disguise $\dfrac{e^{-x}}{e^{-x}}$ and do a bit of algebra:
$\begin{align*} \lim_{x \to \infty} \frac{xe^{x-1}}{(x-1)e^x}&=\lim_{x\to\infty}\left(\frac{xe^{x-1}}{(x-1)e^x}\cdot\frac{e^{-x}}{e^{-x}}\right)\\ &=\lim_{x\to\infty}\frac{xe^{-1}}{x-1}\\ &=\lim_{x\to\infty}\frac{x}{e(x-1)}\\ &=\frac1e\lim_{x\to\infty}\frac{x}{x-1}\\ &=\frac1e\lim_{x\to\infty}\frac{x-1+1}{x-1}\\ &=\frac1e\lim_{x\to\infty}\left(1+\frac1{x-1}\right)\;. \end{align*}$
That last limit really is easy.
You may wonder how I came up with some of the steps. The very first one was simply recognizing that if I divided numerator and denominator by $e^x$, the resulting fraction would be a lot simpler, in that all of the exponentials would be gone. Pulling constant factors outside the limit is usually useful and almost never hurts. The simplification of $\frac{x}{x-1}$ could also have been achieved by doing a straightforward polynomial long division of $x-1$ into $x$, but the trick of subtracting and adding the same amount (here $1$) in order to get an expression that can be split in some nice way is a pretty common one that’s worth remembering.