Which even lattices have a theta series with this property?

I don't have a full answer to your question, and have not tried to work out details, but I can offer some thoughts, at least in even dimension:

• First, to satisfy that property, the theta series needs to be an Eisenstein series which is a Hecke eigenform (using non-trivial constant term and multiplicativity), say with weight $k$, level $N$ and character $\chi$. (Typically $N$ should divide the level $M$ of $\Lambda$ but I think there may be some edge cases to worry about with non-trivial character. At least you have $N | M^2$.)

• Presumably you can use your multiplicativity property to show that your Eisenstein series needs to be new of whatever level $N$ it is.

• Since your theta series must have positive coefficients, the character $\chi$ must be real (so quadratic) and the $p$-th coefficient in the $q$-series should be $p^{k-1} + \chi(p)$. Possibly your multiplicativity condition rules out nontrivial $\chi$.

• Then, as you suggest, you should be able to use facts about Bernoulli numbers, to rule out most weights. E.g., the von-Staudt-Clausen theorem which tells you the denominators should suffice for trivial character.

• If $\Lambda$ is an ideal in a definite quaternion algebra (over $\mathbb Q$ or a totally real field of narrow class number 1) and its (say right) order $O$ is Eichler, then one can show its theta series is an Eisenstein series if and only if we are in class number 1. Namely, the Eichler condition implies there is a unique Eisenstein in the space of theta series of the right $O$-ideals, but this Eisenstein series is a sum over all ideal classes.

Similarly it's reasonable that this (essentially? I'm not familiar with your 16-d example) can only happens for lattices with 1 class per genus.

For half-integral weight forms, I wouldn't expect multiplicativity either, but I don't know where this might be proved. For your case, I would expect you can just check this from details about half-integral weight Eisenstein series, but perhaps someone else can add something more?