Can the commutator of two fundamental groups be a fundamental group?

Let $G=\pi_1(X,x_0)$ and $H=\pi_1(Y,y_0)$. Since $[\pi_1(X),\pi_1(Y)]$ is a subgroup of $\pi_1(X) * \pi_1(Y)$, it is the fundamental group of a certain cover $Z$ of $X\lor Y$, with the quotient $\pi_1(X)\times \pi_1(Y)$ acting on $Z$ by Deck transformations.

We can describe $Z$ as follows. Let $p\colon \bigl(\tilde{X},\tilde{x_0}\bigr)\to (X,x_0)$ and $q\colon\bigl(\tilde{Y},\tilde{y_0}\bigr)\to(Y,y_0)$ be universal covers for $X$ and $Y$, respectively. Then $Z$ is the following subspace of $\tilde{X}\times\tilde{Y}$: $$ \bigl(\tilde{X}\times q^{-1}(y_0)\bigr) \cup \bigl(p^{-1}(x_0) \times \tilde{Y}\bigr). $$ Note that $\pi_1(X)\times \pi_1(Y)$ acts on $\tilde{X}\times\tilde{Y}$ componentwise by Deck transformations, and $Z$ is invariant under this action. It is not hard to see that the quotient of $Z$ by this action is homeomorphic to $X\lor Y$, which proves that it is the desired cover.

For example, if $X$ and $Y$ are circles, then $\tilde{X}=\tilde{Y}=\mathbb{R}$ and $p^{-1}(x_0) = q^{-1}(y_0) = \mathbb{Z}$, so $$ Z \,=\, (\mathbb{R}\times \mathbb{Z}) \cup (\mathbb{Z} \times \mathbb{R}). $$ That is, $Z$ is the standard "grid" in the plane. Note that $Z$ covers $S^1\lor S^1$ in an obvious way, with horizontal lines mapping to the first circle and vertical lines mapping to the second circle.

This is what comes to my mind, but it doesn't address your question exactly. If $\pi_2 X = \pi_2 Y = 0$, then you can take $F$ to be the homotopy fiber of the map $X \vee Y \to X \times Y$. Then you will have a short exact sequence $$0 \to \pi_1 F \to \pi_1 (X \vee Y) \to \pi_1(X \times Y).$$ In general, you will have $$\pi_2(X \times Y) \to \pi_1 F \to \pi_1 (X \vee Y) \to \pi_1(X \times Y).$$ The idea is that "short exact sequences" in spaces are usually taken to "long exact sequences" in algebra.