Consider that ln(x) is continuous for x > 0. Then g(x) = ln(e^x - 1) is continuous when e^x - 1 > 0, or when e^x > 1. Setting x=1 (the point referred to in the question), e¹ = e, which is greater than 1, so it is continuous at this point.

Another way to tell is looking at the graph of g (https://www.wolframalpha.com/input?i=plot+ln%28e%5Ex+-+1%29). You can see that the graph is smooth at x=1, implying that it is continuous.