Suppose $b=a^2$. Note that whenever $a$ is not the identity element $e$, then $a \neq e \Rightarrow a^2 \neq a$, and hence $a$ and $b$ are distinct. Since $a^2 \neq a$, we consider when $a^2 = e$ and otherwise as two cases. Case 1: $a^2 = e$ Then $a$ has a cycle of 2. Axiom (4) is reduced to $a*b = b$, implying that $a=e$, a contradiction. Case 2: $a^2 \neq e$ By Axiom (4), letting $b = a^2$ gives $a^3 * b = b^3 * a^2 \Rightarrow a^5 = a^8 \Rightarrow e = a^3$, by Axiom (2). Therefore, every non-identity element $a$ in $A$ has a cycle of 3 in this case. The property in (4) can be rewritten as $a^3*b = b^3 * a^2 \Rightarrow b = a^2 \Rightarrow a*b = a^3 = e$. So any composition of two distinct, non-identity elements is the identity element $e$ itself. Consider the set generated by a single non-identity element $a$ under the operation $*$. Then the set is $A = \{a, a^2, e\}$, which obeys the four axioms. The addition of another distinct element $b$ gives $a*b = e$ and $a^2*b = e$, but Axiom (2) shows that $a = a^2$, a contradiction. So the maximum cardinality in this case is $\boxed{3}$. Edit: It may be the case that the distinct elements $a,b$ can fall under different cases. We show that a valid set A can only have non-identity elements of the same cycle. From above, an arbitrary element $a \neq e$ can either have a cycle of 2 or 3. Suppose the cycle of $a$ is 2 and the cycle of $b$ is 3. Then Axiom 4 becomes: $a^3*b = b^3*a^2 \Rightarrow a*b = e \Rightarrow a*a*b = a*e \Rightarrow b = a$, which is impossible.

Solved with SnowPanda. The answer is 3, achievable with the set $\{0, 1, 2\}$ and $\ast$ being addition $\pmod 3$. For the rest, assume we have the elements $e$, $a$, and $b$. Let $P(a, b)$ denote the assertion that $a^3\ast b = b^3\ast a^2$. Claim: Either $a = e$, $a^2 = e$, $a^2 = a$, or $a^3 = e$. Proof: Assume none of these are true. Then plugging in $(a, a^2)$ gives $a^5 = a^8$, and cancelling $a$'s from the end gives $a^4 = a$. Then plugging in $(a, a^3)$ -- these can't be equal because then we'd have $a^3 = a^4$, so $a = a^2$ -- gives $a^6 = a^{11}$, so $a = a^6$. Then we get $a^4 = a^6$, so $a = a^3$, and then $a^4 = a^3$ so $a = a^2$, contradiction. Assuming $a\neq e$, we will now eliminate the cases of $a^2 = e$ and $a^2 = a$. For the sake of contradiction, suppose there is some element $a$ satisfying $a^2 = e$. Take $b \neq a, e$, so $P(a, b)$ gives $ab = b^3$, so $a = b^2$. Then $P(b, a)$ gives $b^3a = ab^2 = a^2 = e$. Now $b = ba^2 = b^3a = ab^2 = e$, a contradiction. Next, assume for the sake of contradiction that there is some element $a$ satisfying $a^2 = a$. Take $b\neq a, e$, so $P(a, b)\implies ab = b^3a$. From $P(b, a)$ we have $b^3a = ab^2$, so $ab = ab^2$ and $a = ab$. However, then $a^2 = a = ab = ba \implies a = b$, a contradiction. Assume $a, b \neq e$ with $a^3 = e$ and $b^3 = e$, with $b \neq a, a^2$ (this also means $a \neq b, b^2$, since if $a = b^2$ then $a^2 = b^4 = b$). We must have that $e * a = a$ or $e$, because otherwise $P(e*a, a)$ gives $a = e$. Then $P(a^2, b)$ gets $a^6 * b = b^3 * a^4$, which means $e * b = e * a$. Similarly, $e * x = e * y$ for all $x, y \in \{a, b, a^2, b^2\}$. But then we must have $e * b = e * a = e$, since otherwise we'd have $a = b$ or one of them would be $e$. But then $a * (e * a) = (a * e) * a = a * a$, but $a * (e * a) = a * e = a$ as well, so $a * a = a$ and we return to a previous case. This shows to have at least $3$ elements, each must satisfy $a^3 = e$, and then there can be no elements other than $e$, $a$, and $a^2$.