> How does the definition of semigroup action in terms category theory look like?

Before we can define a semigroup action in terms of category theory, we have to define a semigroup in terms of category theory. A semigroup $S$ may be described as an object $S$ together with a morphism
$\mu:S\times S \rightarrow S$ such that the following diagram in $\mu$ commutes:
$$\begin{matrix}
S\times S\times S & \xrightarrow{1\times\mu} & S\times S \\
\\
\downarrow\ {\mu \times 1} & & \downarrow \mu \\
\\
S\times S & \xrightarrow{\mu} & S
\end{matrix}$$
This means that we have $\mu\circ(1\times\mu)=\mu\circ(\mu\times 1)$, but the commutative diagram is more understandable, because it also show the objects and not just the morphisms.

In this language, a semigroup action may be described as a semigroup $S$, an object $X$ and a morphism $\alpha:S\times X \rightarrow X$ such that the following diagram in $\mu$ and $\alpha$ commutes:
$$\begin{matrix}
S\times S\times X & \xrightarrow{1\times\alpha} & S\times X \\
\\
\downarrow\ {\mu \times 1} & & \downarrow \alpha \\
\\
S\times X & \xrightarrow{\alpha} & X
\end{matrix}$$
This means that we have $\alpha\circ(1\times\alpha)=\alpha\circ(\mu\times 1)$.

> Does this definition make sense when applied to an inverse semigroup?

The next step would be to define an inverse semigroup in terms of category theory. We have to express the identities $x=xx^{-1}x$ and $xx^{-1}yy^{-1}=yy^{-1}xx^{-1}$ in terms of commutative diagrams for this. Hence an inverse semigroup may be described as a semigroup $S$ together with a morphism $\zeta:S\times S\rightarrow S$ such that the following diagrams commute:$$???$$
Well, this is surprisingly difficult, but let's assume that this can be solved. Then we could just apply the definition of semigroup action to this scenario, and we would have a definition. But should we add additional conditions that must be satisfied by an inverse semigroup action? Maybe, but I don't know.

> Does it lead to the intended representation theorem if used with the category of sets and partial function?

Honestly, as long as we don't add additional conditions to be satisfied by a semigroup action, I see no reason why this should lead to the intended representation theorem. And even before that, if we are in a category where partial functions are allowed, we would have to also allow that $\mu$ and $\zeta$ are partial operations, as long as all relevant diagrams are still commutative. And then we are already frankly outside the domain of inverse semigroup theory, before we even began come close to the intended representation theorem.

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This is a partial answer (which may be extended later), but it already indicates that the answer to the question is probably: "No, the proposed solution won't work".