Let $\ell^\star$ be the upper logarithmic density on the set of positive integers, namely $$ \forall X\subseteq \mathbf{N}^+, \,\, \ell^\star(X)=\limsup_n \frac{1}{\ln n}\sum_{x \in X\cap [1,n]}\frac{1}{x}. $$
Problem: Let $X$ be the set of all even positive numbers, and let $Y$ be an arbitrary set of odd positive numbers. Is it necessarily the case that $$ \ell^\star(X \cup Y) = \ell^\star(X)+\ell^\star(Y)? $$
The answer is yes. More in general, if $\alpha \in [-1,\infty[$ and $\mathsf{d}^\ast(\mathfrak F; \alpha)$ is the upper $\alpha$-density (on $\mathbf N^+$) relative to a certain sequence $\mathfrak F = (F_n)_{n \ge 1}$ of nonempty subsets of $\mathbf N^+$ with suitable properties (can make this more precise if requested), viz. the function $$\mathcal P(\mathbf N^+) \to \mathbf R: S \mapsto \limsup_{n \to \infty} \frac{\sum_{i \in S \cap F_n} i^\alpha}{\sum_{i \in F_n} i^\alpha},$$ then $$\mathsf{d}^\ast(\mathfrak F; \alpha)(X \cup Y) = \mathsf{d}^\ast(\mathfrak F; \alpha)(X) + \mathsf{d}^\ast(\mathfrak F; \alpha)(Y)$$ for all disjoint $X,Y \subseteq \mathbf N^+$ such that $$X \in \{S \in \mathcal P(\mathbf N^+): \mathsf{d}^\ast(\mathfrak F; \alpha)(S) + \mathsf{d}^\ast(\mathfrak F; \alpha)(S^c) = 1\}.$$ (I denote by $\mathcal P(\mathbf N^+)$ the power set of $\mathbf N^+$, and let $S^c := \mathbf N^+ \setminus S$ for every $S \subseteq \mathbf N^+$.)
The reason is essentially that if $(a_n)_{n \ge 1}$ and $(b_n)_{n \ge 1}$ are two real sequences for which the limit $\lim_n a_n$ exists and is finite, then $\limsup_n (a_n + b_n) = \lim_n a_n + \limsup_n b_n$.
The OP refers to the case $\alpha = -1$ and $F_n = [\![1,n]\!]$ for every $n$.