The following kind of "nilpotent" construction satisfies the properties you require:

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.

Example 1. The following kind of nilpotent construction satisfies the properties require in the question:

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.

Example 2. We can simply adjust the nilpotent construction from Example 1 to obtain a non-nilpotent example:

Let $E$ and $A$ be as in Example 1, let $G$ be another Banach space and let $K: G \to G$ be any compact operator.

We define $\tilde E = E \times G$ (with the maximum norm) and $\tilde A: \tilde E \times \tilde E \to \tilde E$ as $\tilde A((x,g), \, (y,h)) = (A(x,y),\, K(g+h))$ for $x,y \in E$ and $g,h \in G$. Then:

• $\tilde A$ is not condensing. Indeed, for the set $D \subseteq E$ from Example 1 we have $\tilde A(D \times \{0\}, D \times \{0\}) = A(D,D) \times \{0\}$, so $$ \alpha(\tilde A(D \times \{0\}, D \times \{0\})) = \alpha(A(D,D)) = 2\alpha(D) = 2\alpha(D \times \{0\}). $$

• $\tilde A^2$ is condensing: for all $x,y\in E$ and all $g,h \in G$ we have $\tilde A^2((x,g),\, (y,h)) = (0, 2K^2(g+h))$, so $\tilde A^2(S_1,S_2)$ is compact for all bounded sets $S_1, S_2 \subseteq \tilde E$.

• If $K^2$ is not $0$, then $\tilde A^2 \not= 0$. More generally, if $K$ is not nilpotent, then $A^n$ is non-zero for any $n \in \mathbb{N}$.

Example 3. If we want a still less trivial example, we can construct a space $E$ and a map $A$ with the required properties and such that $A^n(E,E)$ is dense in $E$ for every $n \in \mathbb{N}$. To this end, we redo the construction from Example 1, but with a few modifications.

Again, let $F$ be an infinite dimensional Banach space and set $E = F \times F$ (with the maximum norm). Let $K: F \to F$ be a compact linear operator with dense range (it is easy to construct such an operator if we choose $F$, for instance, to be a classical sequence space such as $\ell^p$ or $c_0$).

Now we define an operator $T: E \to E$ as the operator matrix $$ \begin{pmatrix} K & I_F \\ 0 & K \end{pmatrix}. $$ Then we have $$ T^n = \begin{pmatrix} K^n & nK^{n-1} \\ 0 & K^n \end{pmatrix}. $$ for each integer $n \ge 1$. In particular, $T^n$ is compact for each $n \ge 2$.

Again, we define $A: E \times E \to E$ by $A(x,y) = T(x+y)$ for all $x,y \in E$. Then:

• $A$ is not condensing. To see this, let $D$ be the same set as in Example 1 and let $P: E \to E$ be the projection onto the first component. Then $P$ is contractive, so $\alpha(P(X)) \le \alpha(X)$ for each subset $X$ of $E$. Moreover, we have $$ A(D,D) = T(D) + T(D) \supseteq T(D), $$ so it follows that $$ \alpha(A(D,D)) \ge \alpha(T(D)) \ge \alpha(P(T(D))) = \alpha(B \times \{0\}) = \alpha(D). $$

• For each $n \in \mathbb{N}$ and all $x,y \in E$ we have $A^n(x,y) = nT^n(x+y)$, so $A^n(S_1,S_2)$ is relatively compact for all $n \ge 2$ and all bounded sets $S_1, S_2 \subseteq E$. This shows that $A^n$ is condensing whenever $n \ge 2$.

• We have $A^n(E,E) = T^n(E)$ for each $n \in \mathbb{N}$, so it remains to show that $T^n$ has dense range for each $n$, and to this end it suffices to show that $T$ has dense range. So let $(f,g) \in E = F \times F$. Since $K$ has dense range in $F$; there exists a sequence $(g_k)$ in $F$ such that $Kg_k \to g$. On the other hand, again since $K$ has dense range in $F$, we can find, for each index $k \ge 1$, a vector $f_k \in F$ such that $\|Kf_k + g_k - f\| < 1/k$. Hence, $T(f_k, g_k) \to (f,g)$.