For the sake of illustrating the issue, I am considering the $2\times 2$ tensor $\mathbf{A}=\mathbf{i}\otimes \mathbf{i}+\mathbf{j}\otimes \mathbf{j}$ where vectors $\mathbf{i}$ and $\mathbf{j}$ are two orthogonal unit vectors. Accordingly, in basis $(\mathbf{i},\mathbf{j})$, the matrix of $\mathbf{A}$ reads:
$$[\mathbf{A}]_{\mathbf{i},\mathbf{j}}=\begin{bmatrix} 1 & 0 \cr 0&1\end{bmatrix}$$
The change of basis $\mathbf{e}_1=\mathbf{i}/2$ and $\mathbf{e}_2=\mathbf{j}$ is considered.
- Method 1 (incorrect)
Accordingly, $\mathbf{A}=4\mathbf{e}_1\otimes \mathbf{e}_1+\mathbf{e}_2\otimes \mathbf{e}_2$ and in basis $(\mathbf{e}_1,\mathbf{e}_2)$, the matrix of $\mathbf{A}$ reads:
$$[\mathbf{A}]_{\mathbf{e}_1,\mathbf{e}_2}=\begin{bmatrix} 4 & 0 \cr 0&1\end{bmatrix}$$
- Method 2 (correct)
However, if we conduct a basic change of basis $[\mathbf{A}]_{\mathbf{e}_1,\mathbf{e}_2}=[\mathbf{P}]^{-1}[\mathbf{A}]_{\mathbf{i},\mathbf{j}}[\mathbf{P}]=[\mathbf{A}]_{\mathbf{i},\mathbf{j}}$ where
$$[\mathbf{P}]=\begin{bmatrix} 2 & 0 \cr 0 & 1\end{bmatrix}$$ the above "wrong" result is not retrieved.
What is wrong in Method 1?
What's wrong is that the $P^{-1}AP$ formula is for $(1,1)$-tensors, but you've used it on a $(2,0)$-tensor. They transform differently.
If you want $A$ to be a $(1,1)$-tensor, you need to use the dual basis:
$$A=i\otimes i^*+j\otimes j^*$$
$$=(2e_1)\otimes(\tfrac12e_1^*)+e_2\otimes e_2^*$$
$$=e_1\otimes e_1^*+e_2\otimes e_2^*.$$
By definition, $i^*\cdot i=j^*\cdot j=e_1^*\cdot e_1=e_2^*\cdot e_2=1$, and $i^*\cdot j=j^*\cdot i=e_1^*\cdot e_2=e_2^*\cdot e_1=0$. It follows that, if
$$e_1=\tfrac12i,\;e_2=j,$$
then
$$e_1^*=2i^*,\;e_2^*=j^*.$$