Since $e^x>0$ for all real $x$, then $y=e^x$ does not intersect $y=kx$ at all if $k=0$.
If $k<0$, then observe that $e^x-kx$ is strictly increasing, with $\lim_{x\to\infty}(e^x-kx)=\infty$ and $\lim_{x\to-\infty}(e^x-kx)=-\infty.$ From the strict monotonicity, there is at most one solution to $e^x=kx$, and from an application of the Intermediate Value Theorem, there is at least one solution.
Suppose that $k$ is positive and that $y=kx$ intersects $y=e^x$ at exactly one point--equivalently, that $f(x)=e^x-kx$ has exactly one zero. Now, $f'(x)=e^x-k$, and by observing the sign of $f'(x)$, we conclude that $f$ is decreasing on $(-\infty,\ln k)$ and increasing on $(\ln k,\infty)$, achieving a global minimum when $x=\ln k$. Noting that $f(x)\to\infty$ as $x\to\pm\infty$, it follows that the minimum value of $f(x)$ cannot be negative, for otherwise, $f(x)$ would have two zeroes--one in $(-\infty,\ln k)$ and one in $(\ln k,\infty)$--but on the other hand, the minimum value of $f(x)$ cannot be positive, either, for otherwise $f(x)$ would have no zeroes. Thus, the minimum value of $f(x)$ (which, recall, is achieved at $x=\ln k$) is $0$, meaning $0=f(\ln k)=e^{\ln k}-k\ln k=k-k\ln k=k(1-\ln k).$ Since $k>0$, this means that $1-\ln k=0$, so $\ln k=1$, and so $k=e$.