Is the identity axiom in the definition of group action redundant?

In your "more direct way", the problem is that you can't simplify $g$ without knowing that $e\cdot x=x$ in the first place.

In fact, if your set $X$ is non-empty, then you can always choose some $x_0\in X$ and define an "action" as $g \cdot x=x_0$ for all $x\in X$ and $g\in G$. This satisfies the second condition, but not the first one.

But you are right that any group homomorphism preserve the identity. The problem here is that in general, a map $G\times X\to X$ only gives you a function from $G$ to the monoid $\operatorname{End}(X)$ of maps $X\to X$, and the second condition only tells you that this map preserve the products of these monoids. But for monoids, it is not true that any map that preserve the product also preserve the identity, so you need to ask for that to hold as well, i.e. you need to add the first condition.

If you multiply $g.x = g.(e.x)$ on the left by $g^{-1}$, then you get $g^{-1}.g.x = g^{-1}.g.(e.x)$, which you can simplify to $e.x = e.(e.x)$, but it carries no more information than $e.x = e.x$.

Fix some $y\in X$ and define $g.x=y$ for every $g\in G$ and every $x\in X$.

Then $(gh).x=y=g.(h.x)$ for $g,h\in G$ and $x\in X$ but not $e.x=x$ for every $x\in X$ (unless $X$ is a singleton).