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.