The following kind of "nilpotent" construction satisfies the properties you require:
Let $F$ be an infinite dimensional Banach space and let $E = F \times F$ (say, with the maximum norm). Let $I_F: F \to F$ denote the identity operator and define $T: E \to E$ as the operator matrix $$ \begin{pmatrix} 0 & I_F \\ 0 & 0 \end{pmatrix}. $$ Note that $T^2 = 0$.
Finally, we define $A: E \times E \to E$ by the formula $A(x,y) = T(x+y)$ for all $x,y \in E$. Then:
$A^2(x,y) = A(T(x+y), T(x+y)) = 2T^2(x+y) = 0$ for all $x,y \in E$. In particular, $A^2$ is condensing.
On the other hand, $A$ is not condensing: Let $B \subseteq F$ denote the closed unit ball in $F$ and let $D := \{0\} \times B \subseteq E$. Then $D$ is not compact, so $\alpha(D) > 0$. However, $$ A(D,D) = T(D) + T(D) = B \times \{0\} + B \times \{0\} = 2(B \times \{0\}), $$ so $\alpha(A(D,D)) = 2\alpha(B \times \{0\}) = 2\alpha(\{0\} \times B) = 2\alpha(D) \ge \alpha(D)$.
Remark. The equality $\alpha(B \times \{0\}) = \alpha(\{0\} \times B)$ used above follows from the fact that $E \ni (f,g) \mapsto (g,f) \in E$ is an isometry.