We have $L \cong A^{\oplus \mathbb{N}} \oplus B^{\oplus \mathbb{N}}$ and $A \oplus A^{\mathbb{N}} \cong A^{\mathbb{N}}$, so $A \oplus L \cong L$.
We have $L \cong A^{\oplus \mathbb{N}} \oplus B^{\oplus \mathbb{N}}$ and $A \oplus A^{\mathbb{N}} \cong A^{\mathbb{N}}$, so $A \oplus L \cong L$.