Surreal arithmetic with $\frac{1}{2}\omega$

Your calculation looks right. What you are overlooking is a comment in the earlier chapter which discusses various tricks that help one calculate values. Conway notes that one can often remove elements from Left and/or Right which do not affect the value.

For example:

The Number $$\langle 0, 1, 2, 3 | \rangle$$ is equal to the number $$\langle 3 | \rangle$$ because $0$, $1$, and $2$ are to the left of $3$ and don't affect the value (which is 'between Left and Right').

By the same token, one can remove $\frac12$, $\frac32$ ,$2$, etc. from Left above because there is always an integer greater than the removed ones to force the remainder to be 'to the left of the value'; similarly for Right above.

Pat Muchmore had the right idea. For simplicity I will use $\psi:=\frac{1}{2}\omega$. You calculated $$\psi+\psi=\langle\{1+\psi,2+\psi,3+\psi,\ldots\}|\{\psi+\omega-1,\psi+\omega-2,\psi+\omega-3,\ldots\}\rangle$$ and want to show $\psi+\psi=\omega$. For this, it suffices to show $\psi+\psi\leq\omega$ and $\psi+\psi\geq\omega$.

By definition$^1$, $x\geq y$ iff there aren't any $v\in X^R,w\in Y^L$ with $v\leq y$ or $x\leq w$.

Now for any $n\in\mathbb{N}$ we have $\psi+\omega-n>\omega$, so there is no $x\in(\psi+\psi)^R$ with $x\leq\omega$ (note that $\psi+\psi, \omega\notin (\psi+\psi)^R$). On the other hand, $\psi>n$, then $\psi+\psi>n$ and so there is no $x\in\omega^L$ with $\psi+\psi\leq x$. We conclude $\psi+\psi\geq\omega$.

There isn't any $x\in\omega^R$ with $x\leq\psi+\psi$ because $\omega^R=\{\}$. And for any $n\in\mathbb{N}$ we have $\omega>n+\psi$, therefore there is no $x\in(\psi+\psi)^L$ with $\omega\leq x$. Therefore $\omega\geq\psi+\psi$ and we conclude $$\frac{1}{2}\omega+\frac{1}{2}\omega=\omega$$

$^1$ I'm working with "On numbers and games" from J.Conway, and don't have Knuth's book at hand, but I'm sure $\leq$ is somehow equivalent defined there.