**Edit.** I begin with a **positive** result, which might be what the OP had in mind (although it is not what the OP wrote).

Let $S$ be an algebraic space. Let $f:\mathcal{X}\to S$ and $g:\mathcal{Y}\to S$ be locally finite type algebraic stacks over $S$. For every field $k$ and for every morphism, $$s:\text{Spec}\ k \to S,$$ denote by $\mathcal{X}_s$, resp. by $\mathcal{Y}_s$, the $2$-fiber product $\mathcal{X}\times_S \text{Spec}\ k$, resp. $\mathcal{Y}\times_S \text{Spec}\ k$. Denote by $\mathcal{X}_s\times_{\text{Spec}\ k} \mathcal{Y}_s$ the $2$-fiber product. Each of these is a locally finite type algebraic stack over $\text{Spec}\ k$. For every algebraic stack $\mathcal{Z}$ over $\text{Spec}\ k$, denote by $|\mathcal{Z}(k)|$ the set of $2$-equivalence classes of $1$-morphisms $\text{Spec}\ k \to \mathcal{Z}$ that are sections of the natural morphism $\mathcal{Z}\to \text{Spec}\ k$. Then there is an induced map, $$\phi_s:|\mathcal{X}_s\times_{\text{Spec}\ k}\mathcal{Y}_s(k)| \to |\mathcal{X}_s(k)|\times |\mathcal{Y}_s(k)|.$$

**Proposition.** If $k$ is algebraically closed, then $\phi_s$ is a bijection.

**Proof.** There is a result in Laumon and Moret-Bailly (I believe it is Théorème 11.5, p. 95) that says that stacks are, up to nonreduced structure, unions of locally closed substacks that are gerbes over coarse moduli spaces.

MR1771927 (2001f:14006)

Laumon, Gérard; Moret-Bailly, Laurent

Champs algébriques. (French)

Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge.

A Series of Modern Surveys in Mathematics, 39.

Springer-Verlag, Berlin, 2000. xii+208 pp.

ISBN: 3-540-65761-4

Thus, it suffices to consider the case when $\mathcal{X}$, resp. $\mathcal{Y}$, is a gerbe over a finite type $k$-scheme $X$, resp. is a gerbe over a finite type $k$-scheme $Y$. Since $k$ is algebraically closed, the following induced map is a bijection, $$(X\times_{\text{Spec}\ k}Y)(k) \to X(k)\times Y(k).$$ Thus, finally, we are reduced to the case that $\mathcal{X}$ and $\mathcal{Y}$ are gerbes over $\text{Spec}\ k$. Again, since $k$ is algebraically closed, each of these is of the form $BG$, resp. $BH$, for a finite $k$-group scheme $G$, resp. $H$. Then $BG\times_{\text{Spec}\ k} BH$ equals $B(G\times_{\text{Spec}\ k} H)$. Each of $|BG(k)|$, $|BH(k)|$, and $|B(G\times_{\text{Spec}\ k} H)(k)|$ is a singleton set. Thus, the induced map is a map of singleton sets, and thus it is a bijection. **QED**

**Negative result.** The positive result above typically fails if $k$ is not an algebraically closed field. For instance, for a prime power integer, $q_0=p^e$, for the finite field $\mathbb{F}_{q_0}$, for every $\mathbb{F}_{q_0}$-scheme $X$ that is finite type and **geometrically integral** of dimension $d\geq 1$, by the **Lang-Weil Estimates**, there exists a real number $C>0$ such that for every finite field extension $\mathbb{F}_q/\mathbb{F}_{q_0}$, $$| \#X(\mathbb{F}_q) - q^d| \leq C\cdot q^{d-(1/2)}.$$ Since the exponential function grows more rapidly than polynomial functions, there exists an integer $r_0=r_0(d,C) = r_0(X)$ such that for every $r\geq r_0$, $$q_0^{r/2} > \frac{(1+C)}{2}r\log_2(r) + C.$$ Thus, for every $r\geq r_0$, the set $X(\mathbb{F}_{q_0^r})$ is strictly larger than the union over all integer divisors $s$ of $r$ of the image of the natural set map, $$X(\mathbb{F}_{q_0^s}) \times \text{Hom}_{\mathbb{F}_{q_0}-\text{alg}}(\mathbb{F}_{q_0^s},\mathbb{F}_{q_0^r})\to X(\mathbb{F}_{q_0^r}).$$ Thus, there exists a closed point of $X$ whose residue field is isomorphic to $\mathbb{F}_{q_0^r}$.

For every pair of $\mathbb{F}_{q_0}$-schemes $X$ and $Y$ that are each of finite type and geometrically integral of dimension $\geq 1$, for $r_0 = \max(r_0(X),r_0(Y))$, for every integer $r\geq r_0$, there exists closed points $x\in X$ and $y\in Y$ with residue fields isomorphic to $\mathbb{F}_{q_0^r}$. The set of closed points of $X\times_{\text{Spec}\ \mathbb{F}_{q_0}}Y$ with residue field isomorphic to $\mathbb{F}_{q_0^r}$ and projecting to $x$, resp. $y$, via the first projection, resp. second projection, is isomorphic to the underlying point set of $\text{Spec}\ \mathbb{F}_{q_0^r}\otimes_{\mathbb{F}_{q_0}}\mathbb{F}_{q_0^r}$. Since $\mathbb{F}_{q_0^r}/\mathbb{F}_{q_0}$ is Galois of degree $r$, this point set has size $r$. Thus, the natural map of underlying point sets, $$|X\times_{\text{Spec}\ \mathbb{F}_{q_0}}Y| \to |X| \times |Y|,$$ is not injective.

Next, let $\mathcal{X}$, resp. $\mathcal{Y}$, be an algebraic stack over $\mathbb{F}_{q_0}$ that is of finite type and that contains a dense open substack $X$, resp. $Y$, that is a geometrically irreducible scheme. Since the map of point sets for $X$ and $Y$ is not injective, the same holds for the map of point sets for $\overline{X}$ and $\overline{Y}$.

**Corollary.** For every finite field $\mathbb{F}_{q_0}$,
for $(g_1,n_1)$ and $(g_1,n_2)$ pairs of integers with $g_i>2$, with $g_i=2$ and $n_i>0$, with $g_i=1$ and $n_i > 2$, or with $g_i=0$ and $n_i>4$, then the stacks $\overline{\mathcal{M}}_{g_i,n_i},$ $i=1,2$, each have a dense open substack that is a scheme. Thus, the map of point sets for $\overline{\mathcal{M}}_{g_1,n_1}$ and $\overline{\mathcal{M}}_{g_2,n_2}$ is not injective.

bijectionwhen $X$ is a $k$-scheme of dimension $\geq 1$. There are many points of $|X \times_k X|$ mapping to the generic points of both factors. Surely the situation for stacks is no different. $\endgroup$ – R. van Dobben de Bruyn Dec 24 '17 at 18:41