Group objects in $\textbf{Cat}$ are strict monoidal categories with an antipode functor endofunctor $\text{inv}$ such that the standard diagram of groups (with the appropriate replacements), shown below, commutes.
The following definition is taken from Wikipedia.
A strict 2-group is a strict monoidal category in which every morphism is invertible and every object has a strict inverse (so that $X\otimes Y$ and $Y\otimes X$ are actually equal to the unit object).
It seems to be folklore that strict 2-groups are precisely group objects in $\textbf{Cat}$. Indeed, from the diagram above it is clear that $X\otimes \text{inv}(x) = \text{inv}(x)\otimes X = I$. On the other hand, I can't see why would morphisms be invertible.
Remarkably, this question is avoided in the category of groupoids, i.e. strict 2-groups are clearly group objects in the category of groupoids.
PS: a question very similar to this one has already been posted, but it looks for tensor inverses instead of true inverses.
The following proof also works for non-strict 2-groups (here, "2-group" is used in the "categorized group" sense, not the "$p$-group" sense).
Let $C$ be a (not necessarily strict) monoidal category such that there is a functor $\mathrm{inv}:C \to C$ for which the functors $X \mapsto X \otimes \mathrm{inv}(X)$ and $X \mapsto \mathrm{inv}(X) \otimes X$ are naturally isomorphic to the constant functor with value $I$.
Then, for any morphism $f:X \to Y$ in $C$, the isomorphism $f \otimes \mathrm{inv}(f):X \otimes \mathrm{inv}(X) \to Y \otimes \mathrm{inv}(Y)$ may be factorized in at least two ways, namely through $X \otimes \mathrm{inv}(Y)$ as $(f \otimes 1_{\mathrm{inv}(Y)}) \circ (1_X \otimes \mathrm{inv}(f))$, or through $Y \otimes \mathrm{inv}(X)$ as $(1_Y \otimes \mathrm{inv}(f)) \circ (f \otimes 1_{\mathrm{inv}(X)})$. This shows that $f \otimes 1_{\mathrm{inv}(Y)}$ is a split epimorphism, while $f \otimes 1_{\mathrm{inv}(X)}$ is a split monomorphism.
Also, the functors $- \otimes \mathrm{inv}(Y)$ and $- \otimes \mathrm{inv}(X)$ are equivalences, and so they reflect split epimorphisms and split monomorphisms. It follows then that $f$ must be a split epimorphism and a split monomorphism, so it must in fact be an isomorphism.
Hence, $C$ must in fact be a groupoid. $\square$