It happens quite often that the image of
$$ [ \cdot, \cdot]: \mathfrak{g} \times \mathfrak{g} \rightarrow \mathfrak{g}$$
is a proper subset of $\mathfrak{g}$. An extreme case is given by commutative (also called abelian) Lie algebras, where the Lie bracket is constant $=0$, in other words the above image is just $\lbrace 0 \rbrace$ (but $\mathfrak{g}$ can be any vector space you like).
Other examples are all nilpotent and even all solvable Lie algebras.
It is worth noting that in general, the image of the Lie bracket is not even a vector subspace (scalar multiples are obviously fine, but the sum of commutators has no reason to be a commutator itself). The case that the space generated by the image (which by abuse of notation is usually denoted $[\mathfrak{g}, \mathfrak{g}]$, whereas formally it should be something like "$\langle[\mathfrak{g}, \mathfrak{g}]\rangle$") is all of $\mathfrak{g}$ has a special name: Such a Lie algebra is called perfect.
Now note that your condition is even more restrictive than being perfect! (For examples of perfect Lie algebras where not every element is a commutator, cf. YCor's answer here: https://math.stackexchange.com/a/2365525/96384.)
As Tobias Kildetoft points out in a comment, on the other hand, a big class of Lie algebras for which indeed each element is a commutator are split semisimple Lie algebras, at least if the cardinality of your ground field is big enough (certainly over all infinite fields, and since you say you're a physicist, I assume you are interested only in $\Bbb R$ or $\Bbb C$ anyway). This is a non-trivial theorem. See the reference in Dietrich Burde's answer here: https://math.stackexchange.com/a/771802/96384. For an elementary but by no means trivial proof just for the case of $\mathfrak{sl}_n(\Bbb R)$ and $\mathfrak{sl}_n(\Bbb C)$, see user1551's answer here: https://math.stackexchange.com/a/252324/96384
