Let $G$ be a Lie group.
Suppose $G$ admits a a bi-invariant volume form. Is it true that $G$ admits a bi-invariant (Riemannian) metric?
Since we know that $G$ admits a bi-invariant metric if and only if it is a cartesian product of a compact group and a vector space $\mathbb{R}^n$, this question is equivalent to the following:
Are there any Lie groups which are not products (in the form mentioned above), which admit a bi-invariant volume?
The existence of a biinvariant volume form does not imply the existence of a biinvariant Riemannian metric. For instance, each noncompact connected semisimple Lie group $G$ admits such a form but never a biinvariant metric (the latter part I will skip).
Here is a self-contained argument. Start with a left-invariant Riemannian metric on a connected semisimple Lie group $G$, e.g. $G=PSL(n, {\mathbb R})$. This metric defines a left-invariant volume form $\omega$ on $G$. Now, let us prove that $\omega$ is also right-invariant. Given $g\in G$, consider its adjoint representation $Ad_g$ on ${\mathfrak g}$. Let $det: GL({\mathfrak g})\to {\mathbb R}^*$ be the determinant defined by the form $\omega$ at $T_{1}G={\mathfrak g}$. Then the composition $det\circ Ad$ is a homomorphism $G\to {\mathbb R}^*$, a character. Since $G$ is connected and semisimple, it has only trivial characters, hence $Ad$ preserves the volume form $\omega$ at $1\in G$. From this and left-invariance of $\omega$, you see that $\omega$ is right-invariant.