Yes, and no. There is a "categorically comprehensive" reason for this trace map to exist, but not necessarily to construct it. And to prove that this reason is valid, one does require categorical machinery. The elevator speech answer is:
THH is an algebra in some symmetric monoidal category. K is the unit in this symmetric monoidal category, so THH receives a unique algebra map from K theory.
(If you are looking for a less theoretical reason, and want some explicit constructions, you might take a look at Kantorovitz-Miller, "An explicit description of the Dennis trace map.")
Everything I write below, I learned from Blumberg-Gepner-Tabuada, "Uniqueness of the multiplicative cyclotomic trace."
First, we note that both THH and K define functors
$\infty Cat^{perf} \to Spectra$.
On the righthand side is the category of spectra. (Take any model you like, so long as it's not the homotopy category of the model. You can take Lurie's oo-categorical model, or symmetric spectra if you like.)
On the lefthand side is the category of perfect stable oo-categories. Roughly, these are the categories that look like modules over some ring spectrum. A different way you might describe this category is as follows: The category of spectrally enriched categories, localized with respect to Morita equivalence.
Note that both categories--$\infty Cat^{perf}$ and $Spectra$--have a symmetric monoidal structure. The latter has the usual smash product, while the former has the tensor product of stable $\infty$-categories. This is given by the cocompletion of the following naive tensor product: Given two categories $A$ and $B$, the objects of $A \otimes^{naive} B$ are pairs of objects $(a,b)$, and the hom spectrum between $(a,b)$ and $(a',b')$ is given by $hom(a,a') \wedge hom(b,b')$.
Moreover, we note that both THH and K satisfy the following properties:
They are lax monoidal. (In fact, THH is symmetric monoidal.) This means that we have specified natural maps $K(A) \otimes K(B) \to K(A \otimes B)$, but these need not be equivalences.
They are localizing: If we have a short exact sequence of categories $A \to B \to C$, we have a cofibration sequence of spectra $K(A) \to K(B) \to K(C)$, and likewise for THH. The proof of this for THH can be found in Blumberg-Mandell ("Localization theorems in topological Hochschild homology and topological cyclic homology").
Now, consider the category of all functors $\infty Cat^{perf} \to Spectra$ satisfying (2). One can construct a symmetric monoidal structure on this category. And it turns out that any functor further satisfying (1) can be made into an $E_\infty$ algebra in this category, and that K theory is in fact the unit! Since THH satsifies (1) and (2), the corresponding algebra for THH receives a unique algebra map from K theory.
When $A$ is an $E_\infty$ ring, then $THH(A)$ is an $E_\infty$ ring as well; by the algebra map from $K$ to $THH$, one obtains an $E_\infty$ ring map $K(A) \to THH(A)$.
More generally, if $A$ is an $E_n$-algebra, then $K(A) = K(AMod)$ is an $E_{n-1}$ ring. This is because the category of $A$-modules can be given an $E_{n-1}$-structure, and $K$ theory is lax monoidal. Moreover, you can also prove that $THH(A)$ has an $E_{n-1}$ structure as well (you can see this also using factorization homology, for instance). The fact that there is an algebra map $K \to THH$ implies that one also obtains an $E_{n-1}$-algebra map $K(A) \to THH(A)$.
