Prove That $\int_{1}^{e}\frac{1}{t}dt=1$ Without the Use of Logarithms

I've been studying Real Analysis by Jay Cummings, and am working through the exercises on integration. The question is as such:

Define a function $L:(0, \infty)\rightarrow\mathbb{R}$ by $L(x)=\int_{1}^{x}\frac{1}{t}dt$

It then goes on to prove that $L(xy)=L(x)+L(y)$ without the use of the natural logarithm et cetera... However, the final part is interesting:

Prove that $L(x^y)=yL(x)$, and hence deduce that $L(e^x)=x$. After proving the first part, $L(e^x)=xL(e)$ is trivial. However, I'm not entirely sure what they want me to do after that - just assume that $L$ is the natural logarithm and call it a day with $L(e)=1$? If so, that's very boring, but it got me thinking - how would you prove that $\int_{1}^{e}\frac{1}{t}dt=1$ without assuming that it's simply equal to $ln(e)$? And so I constructed a sequence of partitions $P_n$ with all intervals $[x_{i-1}, x_i]$ having length $\frac{e-1}{n}$, and had a look at the upper and lower Darboux sums, getting (for L):

$$\int_{1}^{e}\frac{1}{t}dt = \lim_{n \to \infty}(\sum_{k=1}^{n}\frac{e-1}{n+k(e-1)})$$

When plugged into Wolfram Alpha, the limit equals 1 as desired. I'd like to do this without the use of a calculator, but I am absolutely stuck! Could someone help me out?

Many thanks