I should probably study the classical viewpoint first. I haven't yet and I will eventually but let us stick to $\infty$-operads for this question. I'm following the lecture notes from Hebestreit-Wagner. Let me recall the terminology first.
One can define an $\infty$-operad $\mathbb{A}\mathrm{ssoc} \to \mathbb{\Gamma}^{\mathrm{op}}$ via finite sets with partially defined maps with a total ordering on each fiber. An algebra over an $\infty$-operad $\mathscr{O}$ in a symmetric monoidal $\infty$-category $\mathscr{C}^{\otimes} \to \Gamma^{\mathrm{op}}$ is an $\infty$-operad map $\mathscr{O} \to \mathscr{C}^{\otimes}$ over $\Gamma^{\mathrm{op}}$ which preserves inerts (i.e. cocartesian lifts of inert maps).
Question. I only want to check $\operatorname{Alg}_{\mathbb{A}\mathrm{ssoc}}(\mathbf{Set}^{\times}) \simeq \operatorname{Mon}(\mathbf{Set})$ or $\operatorname{Alg}_{\mathbb{A}\mathrm{ssoc}}(\mathbf{Ab}^{\otimes}) \simeq \mathbf{Ring}$ but somehow I can't get this.
How do you do this in this setting? Or do people rather prefer to use certain models to compute examples like these (in that case, what's the general strategy?)?
We will argue that $\mathrm{Alg}_{\mathbb{A}\mathrm{ssoc}}(\mathcal{C}^\otimes)\simeq\mathrm{Mon}(\mathcal{C}^\otimes)$ for any monoidal $1$-category $\mathcal{C}^\otimes$.
As a first step, we must find the inerts of $\mathbb{A}\mathrm{ssoc}$. Given an inert $\alpha\colon\langle n \rangle\to\langle m\rangle$, every fiber of $\alpha$ is either empty or a singleton, so has a unique total order regardless. There is hence only one lift of $\alpha$ to a map in $\mathbb{A}\mathrm{ssoc}$, and this is a cocartesian lift.
Next, we recall the way a monoidal $1$-category $\mathcal{C}^\otimes$ is turned into an $\infty$-operad. This is spelled out in Example II.11(e) of the linked lecture notes: essentially, the objects of $(\mathcal{C}^\otimes)_n$ are tuples $(n,x_1,\ldots,x_n)$ (and $(\mathcal{C}^\otimes)_0$ has a unique object $(0,I)$, where $I$ is the unit of the monoidal structure), and a map $(n,x_1,\ldots,x_n)\to(m,y_1,\ldots,y_m)$ is a map $\alpha\colon \langle n\rangle \to\langle m\rangle$ in $\Gamma^\mathrm{op}$, together with a choice of total order on all the fibers, and for each $j=1,\ldots,m$ a map $f_j\colon\bigotimes_{i\in\alpha^{-1}(j)}x_i\to y_j$ in $\mathcal{C}$, where this tensor product is taken in the order that we specified on the fibers $\alpha^{-1}(j)$. The map $\mathcal{C}^\otimes\to\Gamma^\mathrm{op}$ is of course the forgetful map $(n,x_1,\ldots,x_n)\to\langle n\rangle$. Given an inert $\alpha\colon \langle n\rangle\to\langle m\rangle$ in $\Gamma^\mathrm{op}$ defined at elements $a_1,\ldots, a_i\in\langle n\rangle$ and $(n,x_1,\ldots,x_n)\in(\mathcal{C}^\otimes)_n$, the corresponding inert lift of $\alpha$ is the map $(n,x_1,\ldots,x_n)\to(m,x_{\alpha(a_1)},\ldots,x_{\alpha(a_i)})$ given by the morphism $\alpha\colon \langle n\rangle\to\langle m\rangle$, the unique choice of total order on all the fibers, and the maps $\mathrm{id}_{x_{\alpha(a_j)}}\colon x_{\alpha(a_j)}\to x_{\alpha(a_j)}$ for all $j=1,\ldots,i$.
An $\infty$-operad morphism $F\colon\mathbb{A}\mathrm{ssoc}\to\mathcal{C}^\otimes$ is hence the following data:
The map $\langle 2\rangle\to\langle 1\rangle, 1\mapsto 1, 2\mapsto 1$ in particular gives us a map $m\colon x\otimes x\to x$, and the unique map $\langle 0\rangle\to\langle 1\rangle$ gives us a map $e\colon I\to x$. All the relations between maps $f$ in $\mathbb{A}\mathrm{ssoc}$ (and the corresponding relations between their images $Ff$ in $\mathcal{C}^\otimes$) do nothing but require the maps $m$ and $e$ to satisfy the usual monoid conditions in $\mathcal{C}$. You can imagine that that is slightly tedious to fully check, so I will leave it at this, and if you want more details, I can edit them in later.
At this point, we have (modulo some details) shown that an $\mathbb{A}\mathrm{ssoc}$-algebra in $\mathcal{C}^\otimes$ is just a usual monoid object. As for why $\mathbb{A}\mathrm{ssoc}$-algebra homomorphisms correspond to monoid homomorphisms: above Theorem II.43 in the lecture notes, such algebra homomorphisms are defined to just be natural transformations of functors that induce identity transformations after postcomposition with the map $\mathcal{C}^\otimes\to\Gamma^\mathrm{op}$. Given two monoids $(x,m,e)$ and $(y,m',e')$ in $\mathcal{C}$ encoded by functor $F,F'\colon\mathbb{A}\mathrm{ssoc}\to\mathcal{C}^\otimes$, such a natural transformation is therefore given for each $n\geq 0$ by a map $\eta_n\colon (n,x,\ldots,x)\to(n,y,\ldots,y)$ in $\mathcal{C}^\otimes$, and these maps $\eta_n$ need to commute with the maps $Ff$ and $F'f$ for morphisms $f$ in $\mathbb{A}\mathrm{ssoc}$. Our description of these maps $Ff$ and $F'f$ above can convince you that $\eta_n$ is fully determined by $\eta_1\colon x\to y$ in $\mathcal{C}$, and in fact $\eta_n$ corresponds to the map $\eta_1\otimes\ldots\otimes\eta_1\colon x\otimes\ldots\otimes x\to y\otimes\ldots\otimes y$, where we have an $n$-fold tensor product. The fact that our maps $\eta_n$ need to commute with $Ff$ and $F'f$ can therefore be seen to be equivalent to asking for $\eta_1\colon x\to y$ to respect the multiplication maps $m$ of $x$ and $m'$ of $y$, and for it to respect the unit maps $e$ of $x$ and $e'$ of $y$. (This is again slightly too tedious to type up now and is easier seen by staring at the situation for a while, but if you want details, I can edit them in later.) Hence $\mathbb{A}\mathrm{ssoc}$-algebra homomorphisms correspond to monoid homomorphisms.
Finally, since all categories in sight are $1$-categories, it is not hard to see that $\mathrm{Alg}_{\mathbb{A}\mathrm{ssoc}}(\mathcal{C}^\otimes)$ will be (equivalent to) a $1$-category again, so we are done: we do not need to check what happens for higher morphisms.