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
Since $A$ is $4\times 4$ and its eigenvalues are $2$, $-2$, $1$, and $-1$, the minimal and characteristic polynomials of $A$ agree and are both equal to $$(t-1)(t+1)(t-2)(t+2) = (t^2-1)(t^2-4) = t^4 - 5t^2 + 4.$$ In particular, by the Cayley-Hamilton Theorem, $$A^4 - 5A^2 + 4I = 0,$$ and therefore $$B+A = A^4 - 5A^2 + 5I + A = (A^4-5A^2+4I) + (A+I) = A+I.$$
Now notice that $\lambda$ is an eigenvalue of $A$ if and only if $\alpha\lambda+\beta$ is an eigenvalue of $\alpha A+\beta I$, to conclude that the eigenvalues of $B+A=A+I$ are $0$, $-1$, $2$, and $3$. Therefore, the trace is $0-1+2+3 = 4$, and the determinant is $0$ (since $A+I$ is not invertible, or since the determinant is the product of the eigenvalues).