= Solution
\b[A <noncommutative> <algebra> may have a <commutative> quotient:] take $A=K\times M_2(K)$, $B=K$, and $\varphi(\lambda,M)=\lambda$. This is a surjective unital homomorphism. Neither requested change of commutativity can occur for an isomorphism, since an isomorphism transfers the identity $ab=ba$ in both directions.
For the matrix classifications, write $E_{ij}$ for a <matrix unit> and $N=E_{12}+E_{23}$. Then $N^2=E_{13}$ and $N^3=0$. The complete square-zero sets and commutativity results are
$$
\begin{array}{c|c|c}
&\text{commutative?}&\{\alpha:\alpha^2=0\}\\\hline
A_1&\text{yes}&\{0\}\\
A_2&\text{yes}&KE_{12}+KE_{13}\\
A_3&\text{yes}&KE_{13}\\
A_4&\text{no}&KE_{12}\\
A_5&\text{yes}&KE_{13}+KE_{23}.
\end{array}
$$
Indeed $A_1$ is diagonal. In $A_2$ and $A_5$ the two off-diagonal generators have all products zero, so multiplication has the form $(x,u)(x',u')=(xx',xu'+x'u)$. The diagonal entries of a square force $x=0$, after which the off-diagonal part squares to zero. In $A_3=K[N]$, a square-zero element first has scalar coefficient zero and then $(yN+zN^2)^2=y^2N^2$, forcing $y=0$. This argument works also in characteristic two. In $A_4$, a square first forces both diagonal parameters to vanish, leaving precisely $KE_{12}$. Noncommutativity follows from $E_{22}E_{12}=0$ but $E_{12}E_{22}=E_{12}$.
The coordinate map $xI+yE_{12}+zE_{13}\mapsto xI+yE_{13}+zE_{23}$ is an isomorphism $A_2\to A_5$, by the common multiplication formula. Every other pair is nonisomorphic: $A_4$ is the only <noncommutative> member, and among the <commutative> members the dimensions of the square-zero sets are respectively zero, two, one, and two. These sets are vector spaces in these examples, and an <algebra isomorphism> must carry one onto the other. Thus \b[the only distinct isomorphic pair is $A_2\cong A_5$].
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