Likely missing some rigour that hopefully someone else can fill in.

Let g(x) = e\^x - ex - e

We note that lim x--> -infinity g(x) = infinity, lim x --> infinity g(x) = infinity.

We can also show that g'(x) = e\^x - e is just e\^x moved down by e units, so it has exactly 1 zero.

With only value satisfying g'(x) = 0 on continuous, differentiable g, g(x) is U shaped.

We can show that there exist some continuous domain on g, such that g(x) < 0. For example, g(0) = -e < 0.

By intermediate value theorem we obtain g(x) has exactly two zeros.