In Jech's Set Theory he defines $$0=\emptyset,\quad 1=\{0\},\quad 2=\{0,1\}$$ The definition of sum on ordinals is $$\alpha+0=\alpha,\quad \alpha+(\beta+1)=(\alpha+\beta)+1,\quad \alpha+\beta = \lim_{\xi\rightarrow\beta}(\alpha+\xi)$$
where the last equation is for $\beta>0$. The limit of an non-decreasing sequence $\langle a_\xi : \xi <\alpha\rangle$ is defined as $\sup\{a_\xi:\xi<\alpha\}$.
How could this setup show that $1+1=2$? We would need to show $$2=\sup\{1+\xi:\xi < 1\}$$ but how do we compare 2 to an arbitrary $1+\xi$ when it begs the question of how we identify $1+\xi$?
Recall the definition $\alpha + 1 = S(\alpha)$. So $$1+1 = S(1) = 1 \cup \{1\} = \{0\} \cup \{1\} = \{0,1\}=2.$$