Euler proved in "De Transformatione Serium in Fractiones Continuas" Reference: The Euler Archive, Index number E593 (On the
Transformation of Infinite Series to Continued Fractions) [Theorem VI, §40 to §42] that
$$s=\cfrac{1}{1+\cfrac{2}{2+\cfrac{3}{3+\cdots }}}=\dfrac{1}{e-1}.$$
Here his an explanation on how he proceeded. He stated that if
$$\cfrac{a}{a+\cfrac{b}{b+\cfrac{c}{c+\cdots }}}=s,$$
then
$$a+\cfrac{a}{a+\cfrac{b}{b+\cfrac{c}{c+\cdots }}}=\dfrac{s}{1-s}.$$
Since, in this case, we have $a=1,b=2,c=3,\ldots $ it follows
$$1+\cfrac{1}{1+\cfrac{2}{2+\cfrac{3}{3+\cdots }}}=\dfrac{1}{e-2}.$$
Edit: Euler proves first how to transform an alternating series of a particular type into a continued fraction and then uses the expansion
$$e^{-1}=1-\dfrac{1}{1}+\dfrac{1}{1\cdot 2}-\dfrac{1}{1\cdot 2\cdot 3}+\ldots
.$$
REFERENCES
The Euler Archive, Index number E593, http://www.math.dartmouth.edu/~euler/
Translation of Leonhard Euler's paper by Daniel W. File, The Ohio State University.