How are the matrices: $$A=\begin{bmatrix}1&0&0\\0&1&0\\0&0&2\end{bmatrix},\qquad B=\begin{bmatrix}1&1&0\\0&1&0\\0&0&2\end{bmatrix}$$ fundamentally different? These are in Jordan form, so are not conjugate. They have the same eigenvalues, eigenvectors, and eigenspaces.
They have the same characteristic polynomial, and the minimal polynomials differ slightly $\mu_A=(t-1)(t-2)$ and $\mu_B=(t-1)^2 (t-2)$.
A matrix is classified uniquely, up to order of Jordan blocks, by the Jordan normal form. Over $\Bbb C$ is it true that the minimal polynomial classifies the matrix up to this ordering?
What other things are fundamentally different about these matrices?
The matrices \begin{align*} J_1 &=\left[\begin{array}{rr|r|r} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \hline 0 & 0 & 1 & 0 \\ \hline 0 & 0 & 0 & 1 \end{array}\right] & J_2 &=\left[\begin{array}{rr|rr} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \hline 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{array}\right] \end{align*} are in Jordan canonical form and satisfy $\chi_{J_1}(t)=\chi_{J_2}(t)=(t-1)^4$. Their minimal polynomials are $$ \mu_{J_1}(t)=\mu_{J_2}(t)=t^2-2\,t+1 $$ This shows that the minimal polynomial does not determine the Jordan canonical form of a matrix.
Now, let's consider your examples \begin{align*} A &= \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{array}\right] & B &=\left[\begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{array}\right] \end{align*} As you observe, these matrices have the same characteristic polynomials. Thus the eigenvalues of these matrices are $\lambda=1$ and $\lambda=2$. Moreover, the algebraic multiplicities of these eigenvalues are $\DeclareMathOperator{am}{am}\am_A(1)=\am_B(1)=2$ and $\am_A(2)=\am_B(2)=1$.
However, the geometric multiplicities are $\DeclareMathOperator{gm}{gm}\gm_A(1)=2\neq 1=\gm_B(1)$ and $\gm_A(2)=\gm_B(2)=1$.