Is the reaction between thiosulfate and hydronium an acid-base reaction?
Two of the three acid-base theories (Brønsted and Lewis) also define an acid-base reaction while the third theory (Arrhenius) does not — it merely defines what acids and bases are.

In the Lewis theory, an acid-base reaction occurs when a lone pair donor interacts with a lone pair acceptor to form a new bond without a simultaneous redox reaction happening. The less complicated Brønsted-Lowry theory describes an acid-base reaction as a reaction in which a proton $\ce{H+}$ is transferred from an acid to a base, giving us a conjugate base and a conjugate acid, respectively.

As has been pointed out, your reaction is not a one-step reaction. Rather, we have to break it down into its constituent steps. The first step, a reaction of hydronium $\ce{H3O+}$ with thiosulfate $\ce{S2O3^2-}$ has to be an acid-base reaction because that is practically the only reaction hydronium undergoes.

$$\ce{2H3O+ + S2O3^2- -&gt; H2S2O3 + 2 H2O}\tag{1}$$

The acid is the species that donates the transferred proton, the base is the species that accepts it. Therefore, we can assign the acid and base roles as shown in equation $(1')$.

$$\ce{2 \underset{acid}{H3O+} + \underset{base}{S2O3^2-} -&gt; 2 \underset{conjugate base}{H2O} + \underset{conjugate acid}{H2S2O3}}\tag{$1'$}$$

The second reaction is a redox reaction. Thiosulfate (and therefore also its conjugate acid) has two sulfur atoms in the oxidation states of $\mathrm{+V}$ and $\mathrm{-I}$, respectively. These undergo a redox reaction leading to compounds of the oxidation state $\mathrm{+IV}$ (sulfurous acid $\ce{H2SO3}$) and $\pm0$ (elemental sulfur). We can use the full redox reaction determination algorithm, but I’m skipping that since it is a simple one-electron transfer.

$$\ce{H2S2O3 -&gt; H2SO3 + 1/8 S8}\tag{2}$$

And finally, we have another decomposition namely that of sulfurous acid to sulfur dioxide $\ce{SO2}$ and water. This can be seen as the back-reaction of a Lewis acid-base reaction wherein water is the liberated Lewis base and sulfur dioxide the liberated Lewis acid. Indeed, the forward Lewis acid-base reaction (the reverse direction of this reaction) occurs, too. This reaction is in equilibrium, driven to the product side because of the volatility of $\ce{SO2}$.

$$\ce{H2SO3 (aq) &lt;=&gt;&gt; H2O (l)  + SO2 (g) ^}\tag{3}$$

Therefore, the overall reaction is the sum of one Brønsted acid-base reaction, one redox reaction, and one Lewis acid-base reverse reaction.

$$\ce{2 H3O+ (aq) + S2O3^2- (aq) -&gt; 3 H2O (l) + 1/8 S8 (s) + SO2 (g) ^}\tag{4}$$