Consider the following matrix: $$A=\left[\begin{array}{ll} 1 & 1 \\ 4 & 1 \end{array}\right]$$ We need to find $e^{At}$.
Method $1.$ Th eigen values of $A$ are $3,-1$.
I have Diagonalized the matrix as: $$A=PDP^{-1}$$ where $P$ is a matrix with first column as an eigen vector corresponding to eigen value $3$ and second column as eigen vector corresponding to eigen value $-1$. We have: $$P=\left[\begin{array}{cc} 1 & 1 \\ 2 & -2 \end{array}\right], D=\left[\begin{array}{cc} 3 & 0 \\ 0 & -1 \end{array}\right]$$
Now $$e^{At}=Pe^{Dt}P^{-1}=\left[\begin{array}{cc} 1 & 1 \\ 2 & -2 \end{array}\right]\left[\begin{array}{cc} e^{3 t} & 0 \\ 0 & e^{-t} \end{array}\right]\left[\begin{array}{cc} \frac{1}{2} & \frac{1}{4} \\ \frac{1}{2} & \frac{-1}{4} \end{array}\right]=\left[\begin{array}{cc} \frac{e^{3 t}+e^{-t}}{2} & \frac{e^{3 t}-e^{-t}}{4} \\ e^{3 t}-e^{-t} & \frac{e^{3 t}+e^{-t}}{2} \end{array}\right]$$
Method $2.$ We can write the matrix $A$ as: $$A=I+B+C$$ where $B,C$ are Nilpotent matrices of index $2$ given by: $$B=\left(\begin{array}{ll} 0 & 0 \\ 4 & 0 \end{array}\right) \text { and } C=\left(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right)$$
So we have $$e^{Bt}=I+Bt, e^{Ct}=I+Ct$$ So we get $$e^{Bt}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\left[\begin{array}{ll} 0 & 0 \\ 4 t & 0 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 4 t & 1 \end{array}\right]$$ Also: $$e^{Ct}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\left[\begin{array}{ll} 0 & t \\ 0 & 0 \end{array}\right]=\left[\begin{array}{ll} 1 & t \\ 0 & 1 \end{array}\right]$$
Finally we have $$e^{At}=e^{It+Bt+Ct}=e^{It}e^{Bt}e^{Ct}=\left[\begin{array}{ll} e^{t} & 0 \\ 0 & e^{t} \end{array}\right]\left[\begin{array}{ll} 1 & 0 \\ 4 t & 1 \end{array}\right]\left[\begin{array}{ll} 1 & t \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} e^{t} & t e^{t} \\ 4 t e^{t} & \left(4 t^{2}+1\right) e^{t} \end{array}\right]$$
So my answers not matching.
It was not true that$$e^{\operatorname{Id}t+Bt+Ct}=e^{\operatorname{Id}t}e^{At}e^{Ct}.$$More generally, it is seldom true that $e^{A_1+A_2}=e^{A_1}e^{A_2}$.
The other method is fine.