The prime ideals of $(\mathbb{N},\cdot,1,0)$
Let $\mathbb{P}$ be the set of prime numbers. There is a bijection$^1$
$$\mathcal{P}(\mathbb{P}) \to \mathrm{Spec}_{\mathbf{CMon}_0}((\mathbb{N},\cdot,1,0)),~ E \mapsto \langle E \rangle,$$
which maps a set of prime numbers to the ideal generated by it. Explicitly, we have $\langle \emptyset \rangle = \{0\}$ and $\langle E \rangle = \bigcup_{p \in E} p\mathbb{N}$ for $E \neq \emptyset$.
Proof. With the explicit description of $\langle E \rangle$ it is easy to see that $\langle E \rangle$ is indeed a prime ideal with $E = \langle E \rangle \cap \mathbb{P}$. Let $I$ be a prime ideal and $E := I \cap \mathbb{P}$. Clearly, $\langle E \rangle \subseteq I$. Conversely, if $n \in I$, w.l.o.g. $n \neq 0$, we may factor $n$ as a product of prime numbers. Since $I$ is prime, one of the prime numbers, say $p$, must be contained in $I$. Then $p \in E$ and $n \in p\mathbb{N}$, so $n \in \langle E \rangle$. $\square$
The homomorphism $(\mathbb{N},\cdot,1,0) \hookrightarrow (\mathbb{Z},\cdot,1,0)$ induces a bijection on the spectra (you only need to check that a prime ideal $I \subseteq \mathbb{Z}$ satisfies $I = (I \cap \mathbb{N}) \cup -(I \cap \mathbb{N}$), which is easy). So we also have a bijection $\mathcal{P}(\mathbb{P}) \to \mathrm{Spec}_{\mathbf{CMon}_0}((\mathbb{Z},\cdot,1,0))$, $E \mapsto \langle E \rangle$.
Direct sums
Here is a more conceptual explanation and generalization.
Let $(M_i)_{i \in \mathbb{N}}$ be a family of commutative monoids (not with zero at this point), written multiplicatively. Their coproduct $\coprod_{i \in I} M_i$ is the "direct sum" $\bigoplus_{i \in I} M_i \subseteq \prod_{i \in I} M_i$ consisting of those tuples which are $1$ almost everywhere. The inclusions $\iota_i : M_i \hookrightarrow \bigoplus_{i \in I} M_i$ induce maps $\mathrm{Spec}_{\mathbf{CMon}}(\bigoplus_{i \in I} M_i) \to \mathrm{Spec}_{\mathbf{CMon}}(M_i)$ and hence a map
$$\alpha : \mathrm{Spec}_{\mathbf{CMon}}(\bigoplus_{i \in I} M_i) \to \prod_{i \in I} \mathrm{Spec}_{\mathbf{CMon}}(M_i).$$
Conversely, given a family of prime ideals $\mathfrak{p}_i \subseteq M_i$, the ideal $\langle \bigcup_{i \in I} \mathfrak{p}_i \rangle = \bigcup_{i \in I} \langle \mathfrak{p}_i \rangle \subseteq \bigoplus_{i \in I} M_i$ is a prime ideal which restricts to the $\mathfrak{p}_i$ along $\iota_i$. If $\mathfrak{p} \subseteq \bigoplus_{i \in I} M_i$ is any prime ideal, consider some element $m = \prod_{i \in I} m_i \in \mathfrak{p}$. Since $\mathfrak{p}$ is prime, we have $m_i \in \mathfrak{p}$ for some $i$, so $m_i \in \mathfrak{p} \cap M_i$, and since $m$ is a multiple of $m_i$, we see $m \in \langle \mathfrak{p} \cap M_i \rangle$. This shows that $\alpha$ is bijective. (Actually, $\alpha$ is an isomorphism of ringed spaces.)
Adjoining a zero
Notice that for every $M \in \mathbf{CMon}$ there is a canonical bijection
$$\mathrm{Spec}_{\mathbf{CMon}_0}(M \cup \{0\}) \to \mathrm{Spec}_{\mathbf{CMon}}(M),~ \mathfrak{p} \mapsto \mathfrak{p} \cap M.$$
The prime ideals of $(\mathbb{N},\cdot,1,0)$ again
The monoid $(\mathbb{N},+,0)$ has exactly two prime ideals, namely $\emptyset$ and $\mathbb{N}^+$. Prime factorization yields an isomorphism
$(\mathbb{N},\cdot,1,0) \cong (\mathbb{N}^+,\cdot,1) \cup \{0\} \cong \bigl(\bigoplus_{p \in \mathbb{P}} (\mathbb{N},+,0) \bigr) \cup \{0\}$. Thus, the previous results show that
$$\mathrm{Spec}_{\mathbf{CMon}_0}((\mathbb{N},\cdot,1,0)) \cong \prod_{p \in \mathbb{P}} \mathrm{Spec}_{\mathbf{CMon}}((\mathbb{N},+,0)) = \prod_{p \in \mathbb{P}} \{\emptyset,\mathbb{N}^+\} \cong \mathcal{P}(\mathbb{P}).$$
Finally, the tensor product
Now we can also compute the prime ideals of $(\mathbb{N},\cdot,1,0) \otimes_{\mathbb{F}_1} (\mathbb{N},\cdot,1,0)$. It is $(\mathbb{N}^+,\cdot,1) \otimes_{\mathbb{F}_0} (\mathbb{N}^+,\cdot,1)$ with an adjoined zero. The tensor product refers to commutative $\mathbb{F}_0$-algebras aka commutative monoids; I will just write $\otimes$ since the object themselves show which category we are in. We have
$$(\mathbb{N}^+,\cdot,1) \otimes (\mathbb{N}^+,\cdot,1) \cong \bigoplus_{p,q \in \mathbb{P}} (\mathbb{N},+,0) \otimes (\mathbb{N},+,0) \cong \bigoplus_{p,q \in \mathbb{P}} (\mathbb{N},+,0)$$
and therefore
$$\mathrm{Spec}_{\mathbf{CMon}}((\mathbb{N},\cdot,1,0) \otimes_{\mathbb{F}_1} (\mathbb{N},\cdot,1,0)) \cong \mathrm{Spec}_{\mathbf{CMon}}((\mathbb{N}^+,\cdot,1) \otimes (\mathbb{N}^+,\cdot,1)) \cong \mathcal{P}(\mathbb{P} \times \mathbb{P}).$$
Explicitly, the prime ideal associated to a subset $E \subseteq \mathbb{P} \times \mathbb{P}$ is $\bigcup_{(p,q) \in E} \langle p \otimes 1, 1 \otimes q \rangle$ (I think).
$^1$It is a really good idea to not forget forgetful functors in general, especially here when we need to emphasize the multiplicative structure and the zero. This is why I don't just write $\mathbb{N}$, which is merely the underlying set of the monoid with zero $(\mathbb{N},\cdot,1,0)$.
$^2$The term "binoid" is a bad choice here, I won't use it.
Best Answer
Just like in classical algebraic geometry, we have a natural equivalence of spectral Deligne--Mumford stacks $$ \DeclareMathOperator{\Spet}{Spét} \Spet (\mathbf{Z}\underset{\mathbf{S}}{\otimes}\mathbf{Z})\xrightarrow{\simeq} \Spet\mathbf{Z} \underset{\Spet \mathbf{S}}{\times} \Spet \mathbf{Z}, $$ which you can find in Proposition.1.4.11.1.(3) in Lurie's ''Spectral Algebraic Geometry''. The universal property of the Cartesian product on the right is perhaps not super exciting: mapping into this product from an $X$ is just a choice of two maps $f,g\colon X\to \Spet \mathbf{Z}$ (as there is a unique map $X\to \Spet \mathbf{S}$).
As far as I am aware, people think of the homotopy groups of $\mathbf{Z}\otimes_{\mathbf{S}}\mathbf{Z}$ as quite complicated. A hint towards this ''fact'' is that $\mathbf{F}_p\otimes_{\mathbf{S}}\mathbf{F}_p$ is the mod $p$ dual Steenrod algebra. Maybe one could use Bockstein spectral sequences to try and calculate the homotopy groups of $\mathbf{Z}\otimes_{\mathbf{S}}\mathbf{Z}$ from this fact, but even the integral cohomology of things like $K(\mathbf{Z}/2,n)$ is pretty complicated (see Theorem 10.4 of May's ``A general approach to Steenrod operations''). One could also think about using a Tor spectral sequence to calculate $\pi_\ast \mathbf{Z}\otimes_{\mathbf{S}}\mathbf{Z}$, but once one tries to write down an $E_2$-page, one realises this is unfeasible.
However, general stable homotopy theory tells us that $\pi_0 \mathbf{Z}\otimes_{\mathbf{S}}\mathbf{Z}\simeq \mathbf{Z}$, as more generally we have a natural equivalence of abelian groups $$\pi_0 M\underset{R}{\otimes} N\simeq \pi_0 M\underset{\pi_0 R}{\otimes} \pi_0 N$$ for an $E_1$-ring $R$ and left module $N$ and right module $M$, where $R,M$, and $N$ only have homotopy groups in nonnegative degree. We also know that $\Spet R$ has underlying DM-stack $\Spet \pi_0 R$. This says that the underlying classical DM-stack for $\Spet (\mathbf{Z}\underset{\mathbf{S}}{\otimes}\mathbf{Z})$ is just $\Spet\mathbf{Z}$; all the extra information is ``higher'' information.
(This is more of a comment than an answer...I'd also be interested if anyone else had more to add!)