Approach 1
Define $T': P_n \rightarrow P_n$ by $T'(P(x)) = P(x-1)$. We now prove that $T'$ is the inverse of $T$, and therefore $T$ will be an isomorphism (as it has an inverse and is therefore bijective)
Take $P \in P_n$.
$T(T'(P(x))) = T(P(x-1)) = P(x)$
$T'(T(P(x))) = T'(P(x+1)) = P(x)$
Hence, $\forall P \in P_n: T \circ T' = 1_{P_n} = T' \circ T$. By definition, $T' = T^{-1}$
Approach 2
We prove directly that $T$ is bijective.
Let $P \in P_n$
Then $T(P(x-1)) = P(x)$. Hence, $T$ is surjective. Because $P_n$ has finite dimension and the mapping is an endomorphism, it follows that $T$ is bijective, and therefore an isomorphism.