The general rule is correct as you have stated it. This is because there is only one way to add two $l$ angular momenta to make a total angular momentum of $L$; this combination will be either even or odd under exchange of the two sectors. Since the total electron state needs to be exchange-antisymmetric, and the singlet and triplet states are respectively antisymmetric and symmetric, only one of the combinations will be possible.
The key part of the argument is that the combination of $l\oplus l=L$ will always give some definite exchange symmetry. The simplest way I can see of proving this is reducing the couplings to the Wigner $3j$ symbols, which have definite symmetries under exchange (cf. the DLMF here), which specify to
$$
\begin{pmatrix} l&l&L \\ m_2 & m_1 & -M \end{pmatrix}
=
(-1)^{l+l+L}
\begin{pmatrix} l&l&L \\ m_1 & m_2 & -M \end{pmatrix}
$$
i.e. the coupling will have $(-1)^L$ parity under exchange, independently of the value $l$.
This means that you can strengthen your statement to read
the coupling of two electrons from the same shell, $nl^2$, can only form the term $^3L$ if $L$ is odd and the term $^1L$ if $L$ is even.
On the other hand, it is perfectly possible for an atom to exhibit both $^1L$ and $^3L$ terms for any $L$ - you just need to combine electrons from different shells. Thus, you can have both $^1P$ and $^3P$ terms in the $1s2p$ shell of neutral helium, along with a slew of other similar states which you can explore via the NIST energy levels database.
