If you calculate the characteristic polynomial $\chi_A(x)=x^3-x$, you know from Cayley-Hamilton theorem that $P^3-P=0$, i.e. $P^3=P$.

This implies that $P=P^3=P^5=\dots$ and $P^2=P^4=P^6=\dots$. Hence $$ \begin{align} e^{Pt} &= I+Pt + \frac{P^2t^2}2 + \frac{P^3t^3}{3!} + \dots = \\ &= I + P\left(t+\frac{t^3}{3!}+\frac{t^5}{5!}+\dots\right) + P^2\left(\frac{t^2}2+\frac{t^4}{4!}+\frac{t^6}{6!}+\dots\right) = \\ &= I+ P \frac{e^t-e^{-t}}2 + P^2 \left(\frac{e^t+e^{-t}}2-1\right) = \\ &= I+ P\sinh t + P^2(\cosh t-1) \end{align} $$

Using $P=\begin{pmatrix}0&1&0\\0&0&1\\0&1&0\end{pmatrix}$ and $P^2=\begin{pmatrix}0&0&1\\0&1&0\\0&0&1\end{pmatrix}$ I get $$e^{Pt}= \begin{pmatrix} 1&\sinh t&\cosh t-1\\ 0&\cosh t&\sinh t\\ 0&\sinh t&\cosh t \end{pmatrix}.$$

If I did not miss something, this seems to be the same result as WolframAlpha returns. (Up to some algebraic manipulation.)