Does classifying an integer as a discrete log require it be part of a multiplicative group? This question is not a question about the discrete log problem, the generalized discrete log problem, or an additive group. 

The confusion is whether any integer can be considered a discrete log or whether a discrete log has as a precondition, that it be part of a multiplicative group. This wikipedia would seem to indicate that the answer is yes.

For example 0 doesn't have a multiplicative inverse and is therefore not part of a multiplicative group.
The discrete logarithm $\log_b a$ is an integer $x$ such that  $b^x = a$. Similarly to the logarithms, we need a base, here $b$.

If the base is a generator of the group $g$ then any element of the group can be written as a power of the $g$ for some $k$, $y = g^k$. Therefore, the discrete log of $y$ according to base $g$ is $k$.

Take a generator $g$ of a multiplicative group $G$ with order $n$, and then  take $g'=g^k$ where $\gcd(k,n) \neq 1$. Now the $g'$ will generate a subgroup $G'\leqslant G$, not the full group. Then any element of the full group $ a \in G$ and $a \not\in G'$ has not discrete logarithm according to base $g'$, even it is not a member of the subgroup.

When we consider the non-zero elements of a field $F\backslash\{0\}$ they are forming a cyclic group under multiplication. For a proof see the Theorem 1.