$$T:Mat_{m\times m}(F)_F\to Mat_{m\times m}(F)_F:B\mapsto AB$$
$\{E_{ij}=(e_{pq})\in Mat_{m\times m}(F):i,j=1,2,...,m\}$ is a basis of $Mat_{m\times m}(F)$ where
$$e_{pq}=\begin{cases}1\text{ if $i=p,j=q$ ;}\\0\text{ otherwise.}\end{cases}$$
It's a matter of verification that the $i^{th}$ column of $A$ becomes the $j^{th}$ column of
$T(E_{ij})=AE_{ij}.$ Take $A=(a_1, a_2,...,a_m)$ column-wise. Then
$$T(E_{11})=(a_1,0,0,...,0)=a_{11}E_{11}+a_{21}E_{21}+...+a_{m1}E_{m1}\\T(E_{12})=(0,a_1,0,...,0)=a_{11}E_{12}+a_{21}E_{22}+...+a_{m1}E_{m2}\\...\\T(E_{1m})=(0,0,0,...,a_1)=~...\\T(E_{21})=(a_2,0,0,...,0)=~...\\...\\T(E_{2m})=(0,0,0,...,a_2)=~...\\...\\...\\T(E_{m1})=(a_m,0,0,...,0)=~...\\...\\T(E_{mm})=(0,0,0,...,a_m)=~...\\$$
Therefore (verify) $$T=\begin{pmatrix}a_{11}I_{}&a_{12}I&...&a_{1m}I\\a_{21}I&a_{22}I&..&a_{2m}I\\...&...&...&...\\a_{m1}I&a_{m2}I&...&a_{mm}I\end{pmatrix}$$
where $I$ is the identity matrix of order $m.$
This answer assumes the matrices are taken over $\mathbb C$.
Yes, the statement is still true even if the matrix isn't diagonalizable.
For the proof you saw it is sufficient that $D$ can be taken to be an upper triangular matrix (and it can be taken in such a way, this is Schur's Decomposition Theorem). This is enough because its diagonal entries will be the eigenvalues of the starting matrix.
Jordan Canonical Form is also sufficient, but Schur's Decomposition is a weaker condition.
For completeness I'll add the proofs here.
Let $n\in \mathbb N$ and $A\in \mathcal M_n(\mathbb C)$. Let $\lambda _1, \ldots ,\lambda _n$ be the eigenvalues of $A$. The characteristic polynomial $p_A(z)$ of $A$ is $\color{grey}{p_A(z)=}(z-\lambda _1)\ldots (z-\lambda _n)$.
Schur's Decomposition guarantees the existence of an invertible matrix $P$ and an upper triangular matrix $U$ such that $A=PUP^{-1}$ and $U$'s diagonal entries are exactly $\lambda _1, \ldots ,\lambda _n$.
Since similarity preserves the characteristic polynomial, it follows that the characteristic polynomial $p_U(z)$ of $U$ is $\color{grey}{p_U(z)=}(z-\lambda _1)\ldots (z-\lambda _n)$, therefore $U$ and $A$ have the same eigenvalues with the same algebraic multiplicity.
From the fact that $U$'s diagonal entries are $\lambda _1, \ldots ,\lambda _n$ it follows that the trace of $U$ is the sum of the eigenvalues of $A$ and the determinant of $U$ is the product of the eigenvalues of $A$.
Trace properties yield the following $$\text{tr}(A)=\text{tr}\left(PUP^{-1}\right)=\text{tr}\left(UP^{-1}P\right)=\text{tr}(U),$$ thus proving that the sum of the eigenvalues of $A$ equals $\text{tr}(A)$.
Similarly for the determinant it holds that $$\det(A)=\det\left(PUP^{-1}\right)=\det\left(P\right)\det\left(U\right)\det\left(P^{-1}\right)=\det(U),$$
hence the product of teh eigenvalues of $A$ equals the determinant of $A$.
Best Answer
Specifying the determinant and trace doesn't restrict the set of all matrices as much as you'd like it to. In lower dimensions (namely $n=2$), things are a little better, but for $n\geq 3$ the situation isn't all that great.
One thing we do know is that the determinant and trace both show up in the characteristic polynomial in the matrix. For example, in the $2\times 2$ case, if the characteristic polynomial of $A$ factors into $(t-a)(t-b)$, then $$ (t-a)(t-b) = t^2 - (a+b)t + ab = t^2 - \text{trace}~A\cdot t + \det A. $$ At least in this case, having the determinant and trace completely determines the characteristic polynomial, and conversely. Sadly, this is known not to be enough to completely determine a matrix up to similarity. In particular, if the eigenvalues are repeated in the $2\times 2$ case then there are two possible forms of the Jordan matrices. On the other hand, if the eigenvalues are distinct then the matrix is diagonalizable, so in this specific case among $2\times 2$ matrices we can determine $A$ up to similarity explicitly.
In higher dimensions, we can show the same thing happens, but we get additional terms whose coefficients are traces of what are known as exterior powers of $A$. But if the matrix is unkonwn a priori then you won't be able to determine the characteristic polynomial with just the determinant and trace.
Of course, if one knows the Jordan form then one knows essentially everything one could possibly want about the matrix.