It is not true in general that the product of two quotient maps is a quotient maps (I don't know any examples though).
Are any weaker statements true? For example, if $X, Y, Z$ are spaces and $f : X \to Y$ is a quotient map, is it true that $ f \times {\rm id} : X \times Z \to Y \times Z$ is a quotient map?
Your weaker statement is almost true.
If $f: X \to Y$ is a quotient map and $Z$ is locally compact, then $f \times \operatorname{id}$ is a quotient map. I believe that this result is due to Whitehead.
More generally, if $f: X \to Y$ and $g: Z \to W$ are quotient maps and $Y$ and $Z$ are locally compact, then the product $f \times g: X \times Z \to Y \times W$ is a quotient map.
Why? Use the Whitehead theorem twice, since $f \times g = (\operatorname{id} \times g) \circ (f \times \operatorname{id})$.
See Munkres $\S 22$ for counterexamples.
@Ittay: In a cartesian closed category of spaces, a product of identification maps is an identification map. Here is a typical proof essentially from Topology and Groupoids p. 192.
It suffices to prove that if $f:Y \to Z$ is an identification map, then so also is $f \times 1: Y \times X \to Z \times X$ for any $X$.
Let $g:Z \times X \to W$ be a function such that $l=g(f \times 1):Y \times X \to W$ is continuous. By cartesian closedness, we have associated maps
$$ l': Y \to K(X,W), \quad g':Z \to K(X,W)$$
where $K(X,W)$ is the internal hom, and $g'f=l'$. Since $l'$ is continuous, and $f$ is an identification map, then $g'$ is continuous. Hence $g$ is continuous.
If the quotient maps $p : W \rightarrow X, q:Y \rightarrow Z$ are open, then the product map is also a quotient map:
$U\times V$ open in $X \times Y \implies p^{-1}U\times q^{-1}{V} = (p\times q)^{-1}(U\times V)$ open.
Conversely, $U\times V = (p\times q)(\dots) = p\circ p^{-1}U\times q\circ q^{-1}V$ is open since $p,q$ are open maps.
In the category of compactly generated spaces, I think that the product of quotient maps is (always) a quotient map.
There is another example that I just learnt from exercise 16, section 2.2 of Algebraic Topology by Tammo Tom Dieck. This is :
Let $X$ be a topological space and $A$ be a compact subspace of $X$. Denote the quotient map $p:X \to X/A$. Then $ p \times id_Y : X \times Y \to X/A \times Y$ is a quotient map for any arbitrary space $Y$.
@PaulaCartagenaAtará @Jakobian
The proof of this is essentially the tube lemma.
To see this:
Let $L \subseteq X/A \times Y$ be such that $(p \times 1)^{-1}(L) \subseteq X \times Y$ is open.
Pick any $([x_0], y_0) \in L$. We shall construct an open neighborhood of this point that is contained in $L$. This would then imply that $L$ is open exactly when $(p \times 1)^{-1}(L)$ is open.
Case 1: If $[x_0] = A$, then it follows that $A \times \{y_0\} \subseteq (p \times 1)^{-1}(L)$. By the tube lemma, there is an open neighborhood $U$ of $A$ and an open neighborhood $V$ of $y_0$ such that $A \times \{y_0\} \subseteq U \times V \subseteq (p \times 1)^{-1}(L)$.
Now, notice that $p^{-1}(p(U)) = U$ as $A \subseteq U$, so $p(U)$ is open in $X/A$. Thus, $p(U) \times V$ is open in $X/A \times Y$. Thus, $(A, y_0) \in p(U) \times V \subseteq L$ as required.
Case 2: If $[x_0] \neq A$, then write $S:= \{y \in Y : (A, y) \in L\}$.
If $y_0 \notin S$, since $(x_0, y_0) \in L$, there exist $U \subseteq X$, $V \subseteq Y$ both open such that $(x_0, y_0) \in U \times V \subseteq (p \times 1)^{-1}(L)$. As $y_0 \notin S$, it follows that $U \cap A = \emptyset$. This implies that $p^{-1}(p(U)) = U$ and so $p(U)$ open in $X/A$. This gives that $([x_0], y_0) \in p(U) \times V \subseteq L$ and $p(U) \times V$ is open in $L$, as required.
If $y_0 \in S$, then as in case 1, $A \times \{y_0\} \subseteq (p \times 1)^{-1}(L)$. So, by the tube lemma, there is an open neighborhood $U$ of $A$ and an open neighborhood $V$ of $y_0$ such that $A \times \{y\} \subseteq U \times V \subseteq (p \times 1)^{-1}(L)$.
As $x_0 \notin A$, there exist open neighborhoods $U_1$ and $V_1$ of $x_0, y_0$, respectively such that $(x_0, y_0) \in U_1 \times V_1 \subseteq (p \times 1)^{-1}(L)$.
Write $U' := U \cup U_1$ and $V' := V \cap V_1$. Then,
$(x_0, y_0) \in U' \times V' \subseteq (p \times 1)^{-1}(L)$ and also $p^{-1}(p(U')) = U'$ as $A \subseteq U'$. This implies that $p(U') \times V'$ is open in $X/A \times Y$, contained in $L$ and it also contains $([x_0], y_0)$, as required.
The standard theorem is certainly this:
If $f: X \to Y$ is a quotient map and $Z$ is locally compact, then $f \times \operatorname{id}_Z$ is a quotient map.
See Sammy Black's answer.
This theorem considers an arbitrary quotient map $f$ and gives a condition on $Z$ assuring that $f \times \operatorname{id}_Z$ is a quotient map.
However, there is another type of theorem which identifies special quotient maps $f$ such that $f \times \operatorname{id}_Z$ is a quotient map for each space $Z$. An example can be found in user90041's answer. For a proof see ozymandias's answer.
Here is one more example.
Theorem 1. Suppose that $f$ is a quotient map such that $Y$ is locally compact and $f$ is proper (this means that preimages of compact sets are compact). Then $f \times \operatorname{id}_Z$ is a quotient map for any space $Z$.
This is follows from Lemma 4 in
J. H. C. Whitehead: Note on a theorem due to Borsuk, Bull. Amer. Math. Soc. 54 (1948), 1125-1132
Actually Whitehead proves even more.
Theorem 2. Suppose that $f$ is a quotient map such that $Y$ is locally compact and $f$ is proper. Then $f \times g$ is a quotient map for any quotient map $g : Z \to Z'$.
Proof. Let $V \subset W \times Z'$ be a set such that $W = (f \times g)^{-1}(V) \subset X \times Z$ is open. We have to show that $V$ is open.
Let $(y_0,z'_0) \in V$. Pick $(x_0,z_0) \in W$ such that $(f \times g)(x_0,z_0) = (y_0,z'_0)$. Since $f$ is a quotient map, also $f_0 : X \times \{z_0\} \to Y \times \{z'_0\}, f_0(x,z_0) = (f(x),z'_0)$, is a quotient map. Note that $f_0$ is a restriction of $f \times g$.
We have $$(X \times \{z_0\}) \cap W = f_0^{-1}((Y \times \{z'_0\} ) \cap V) . \tag{1}$$ In fact, $(x,z_0) \in f_0^{-1}((Y \times \{z'_0\} ) \cap V)$ iff $f_0(x,z_0) = (f \times g)(x,z_0) \in (Y \times \{z'_0\}) \cap V$ iff $(f \times g)(x,z_0) \in V$ iff $(x,z_0) \in W$ iff $(x,z_0) \in (X \times \{z_0\}) \cap W$.
The set $(X \times \{z_0\}) \cap W$ is open in $X \times \{z_0\}$, thus $(Y \times \{z'_0\}) \cap V$ is open in $Y \times \{z'_0\}$.
Since $Y$ is locally compact and $(y_0,z'_0) \in (Y \times \{z'_0\}) \cap V$, there exists a compact neighborhood $K$ of $y_0$ in $Y$ such that $K \times \{z'_0\} \subset (Y \times \{z'_0\}) \cap V$. The set $f^{-1}(K) \subset X$ is compact and $(1)$ shows $f^{-1}(K) \times \{z_0\} = f_0^{-1}(K \times \{z'_0\} ) \subset W$. Let $$U = \{z \in Z \mid f^{-1}(K) \times \{z\} \subset W \}.\tag{2} $$
Since $W$ is open in $X \times Y$, the tube lemma shows that $U$ is open in $Z$. Clearly $z_0 \in U$. Moreover
$$U = \{ z \in Z \mid K \times \{g(z)\} \subset V\} . \tag{3}$$ In fact, if $f^{-1}(K) \times \{z\} \subset W$, then $K \times \{g(z)\} = (f \times g)(f^{-1}(K) \times \{z\}) \subset (f \times g)(W) = V$. Conversely if $K \times \{g(z)\} \subset V$, then $f^{-1}(K) \times \{z\} \subset (f \times g)^{-1}(K \times \{g(z)\}) \subset (f \times g)^{-1}(V) = W$.
We conclude
$$g(U) = \{ z' \in Z' \mid K \times \{z'\} \subset V\} .\tag{4}$$
The inclusion $\subset$ is trivial. Conversely, let $ K \times \{z'\} \subset V$. Pick $z \in Z$ such that $g(z) = z'$. Then $K \times \{g(z)\} \subset V$, i.e. $z \in U$. Hence $z' = g(z) \in g(U)$.
This implies
$$g^{-1}(g(U)) = U.$$
In fact, $z \in g^{-1}(g(U))$ iff $g(z) \in g(U)$ iff $K \times \{g(z)\} \subset V$ iff $z \in U$.
Since $g$ is a quotient map, $g(U)$ is open in $Z$ and $z'_0 = g(z_0) \in g(U)$. Hence $K \times g(U)$ is a neighborhood of $(y_0, z'_0)$ in $Y \times Z'$. By construction, $K \times g(U) \subset V$.
In above cases, the results follow (easily) from the good behavior of compact open topology (related with locally compact topology, and with the category of compactly generated spaces respectively).