For example if $\lambda=\omega$ then $1+\omega=\omega$
Indeed $1+\omega=\bigcup_{\alpha\in\omega}(1+\alpha)$ but the sum is commutative in $\omega$ so that $1+\omega=\bigcup_{\alpha\in\omega}(\alpha+1)$
Now obviously $\alpha\in(\alpha+1)\subseteq\bigcup_{\alpha\in\omega}(\alpha+1)$ so that $\omega\subseteq 1+\omega$ but if $\beta\in 1+\omega$ then $\beta\in\alpha+1$ so that $\beta$ is a natural number and so that it is an element of $\omega$ and thus $1+\omega\subseteq \omega$.
Finally we conlude that $$ 1+\omega=\omega $$
So how prove that $$ 1+\lambda=\lambda $$ if $\lambda$ is a limit ordinal strictly greater than $\omega$?
In fact if $\lambda\geq\omega$, then $1+\lambda=\lambda$, regardless of whether $\lambda$ is a limit ordinal or not. Let's prove this by induction.
The base case is $\lambda=\omega$, you wrote in the question that you already know that $1+\omega=\omega$, so I won't add anything.
In the successor case suppose that $\lambda=\beta+1$. Then $$1+\lambda=1+(\beta+1)=(1+\beta)+1=\beta+1=\lambda,$$ where $1+\beta=\beta$ holds by inductive hypothesis.
In the limit case we have, by definition, $$1+\lambda=\sup_{\beta<\lambda}1+\beta=\sup_{\beta<\lambda}\beta=\lambda,$$ where $1+\beta=\beta$ for all $\beta<\lambda$ by inductive hypothesis.
Anyway you can prove the result also remembering that if $\omega\le\alpha$ then it exists a unique $\beta$ such that $$ \alpha=\omega+\beta $$ so that if $1+\omega=\omega$ then through the associativity we conclude that $$ 1+\alpha=1+(\omega+\beta)=(1+\omega)+\beta=\omega+\beta=\alpha $$