Proof
Using the definition of conditional probability,
\begin{align*}
P( X > a + b | X > b ) & = P( X > a + b \cap X > b ) \ P( X > b)\\
& = P( X > a + b ) / P( X > b)\\
& = e^{-(a+b)/ \mu} / e^{-b / \mu}\\
& = e^{-a/ \mu}\\
& = P(X > a)
\end{align*}