Consider the (nonsymmetric) positive-definite triadiagonal matrix $P$ of diagonal $2$ and sub-diagonal $-1$, except for the last row which is $[0\ \dots\ 0 \ 1]$ and define
$$ M_n=\begin{bmatrix} 0_n & I_n \\ -P & 0_n \end{bmatrix}.$$
For example, for $n=3$, $$M_3=\begin{bmatrix} 0&0&0&1&0&0\\ 0&0&0&0 &1&0\\ 0&0&0&0&0&1\\ -2&1&0&0&0&0 \\ 1&-2&1&0&0&0\\ 0&0&-1&0&0&0 \end{bmatrix}.$$
Can I simplify $\exp(tM_n)$ for $t\in\mathbb{R}$, i.e. find an appropriate basis which would make the expression simpler?
The characteristic polynomial of $M_3$ is $(1 + X^2)^2 (3 + X^2)$, for $M_4$ it is $(1 + X^2) (2 + X^2) (2 + 4 X^2 + X^4)$... it's does not seem simple.
$P$ has square root $P^{1/2}$ and $M_n$ can be made skew-symmetric:
$$\begin{bmatrix} P^{1/2} & 0_n \\ 0_n & I_n \end{bmatrix} \begin{bmatrix} 0_n & I_n \\ -P & 0_n \end{bmatrix} \begin{bmatrix} P^{1/2} & 0_n \\ 0_n & I_n \end{bmatrix}^{-1} = \begin{bmatrix} 0 & P^{1/2} \\ -P^{1/2} & 0 \end{bmatrix}$$ and the exponential of a skew-symmetric matrix is orthogonal (because if $M=-M^\top$, then $M$ and $M^\top$ commute so $\exp(M+M^\top)=I=\exp(M)\exp(M^\top)$) but that's all what I can find.
You are wrong. Since $P$ is not symmetric, the matrix $B=\begin{pmatrix}0&P^{1/2}\\-P^{1/2}&0\end{pmatrix}$ is not skew-symmetric.
Let $p_n(x)=\det(xI_n-P_n)$ and $q_n(x)=\det(xI_{2n}-M_n)=\det(x^2I_n+P_n)$. Thus, up to a signum, $q_n(x)=p_n(-x^2)$. Now, the $n$ eigenvalues of $P_n$ are $\{1\}\cup \{2+2\cos(\dfrac{k\pi}{n}),k=1,\cdots,n-1\}$.
Clearly, the eigenvalues of $P$ are $>0$ and, consequently, we can define $P^{1/2}$ as a matrix that has only $>0$ eigenvalues. Yet, a priori, $e^B$ is not orthogonal.
Finally $e^{tM}=\begin{pmatrix}\cos(tP^{1/2})&\sin(tP^{1/2})P^{-1/2}\\-\sin(tP^{1/2})P^{1/2}&\cos(tP^{1/2})\end{pmatrix}$.