It is a common convention to have the multiplication/operation implicit when using homomorphisms, or in general in topics in abstract algebra.
More explicitly: take $(G,\ast),(H,\circ)$ as groups. Then $\phi : G \to H$ is a homomorphism if and only if, $\forall a,b \in G$,
$$\phi(a \ast b) = \phi(a) \circ \phi(b)$$
But this notation can be a bit cumbersome sometimes. And notice something else: $G,H$ only have one operation defined on each. So if we were to define $ab$, sans the $\ast$, there's only really one way to interpret it that makes mathematical sense. Similarly, $\phi(a)\phi(b) \equiv \phi(a) \circ \phi(b)$, because there is no other sensible way to interpret that multiplication.
Be it isomorphisms or homomorphisms, this inherently stays the same: the multiplication implied is that of whatever group you lie in. Indeed, many times $\ast$ in one group will be a lot different from $\circ$ in another, even if the two are isomorphic (at least on a surface level, because, remember, "isomorphic" just means that "for all intents and purposes in this theory, the two items are functionally identical").
I think that the problem is simply a flaw in the formulation of your definition and of your claim.
The phrase $G$ acts on $A$ is not well-defined; it should only be used in a very clear context where the action itself is already given, either explicitly or implicitly.
What do I mean by the action itself? I mean that map $G \times A \to A$ in your definition. Let me reformulate your definition:
Definition Let $G$ be a group and $A$ a set. An action of $G$ on $A$ is a map $G \times A \to A$ denoted by $g*a$ satisfying... [now copy the rest of the definition as stated].
Your claim is similarly flawed. I would suggest breaking it into two separate parts.
Claim Let $G$ be a group and $A$ a set.
- For any action $G \times A \to A$ denoted by $g*a$ the map $\varphi : G \to S_A$ defined by $g \mapsto \sigma_g$, where $\sigma_g(a)=g*a$, is a group homomorphism.
- For any group homomorphism $\varphi : G \to S_A$ the map $G \times A \to A$, defined by $(g,a) \mapsto \varphi(g)(a)$, is an action of $G$ on $A$.
And with that done, your proof can be similarly fixed, and you might even allow yourself to use that tricky phrase $G$ acts on $A$ if the context is clear.
Best Answer
An endomorphism of $(G,*)$ is, by definition, a homomorphism from $(G,*)$ to $(G,*)$. Thus, in your formulation, we require $* = \cdot$ (as functions $G\times G \to G$).