I will reuse the same trick as in the answer to your other question to show that the answer to this is still in the negative.
Let $f$ and $g$ be two upper densities (in the sense of the OP), and let $\alpha \in [0,1]$ and $q \in [1,\infty[$. Then the function $$h := ((1-\alpha) f^q + \alpha g^q)^{\frac{1}{q}}$$ is an upper density too (most notably, condition (F3) follows from Minkowski's inequality, which is where we need $q \ge 1$).
In particular, we will assume from now on that $f$ is the upper asymptotic (or natural) density (on $\mathbf N^+$), namely the function $$ \mathcal P(\mathbf N^+) \to \mathbf R: X \mapsto \limsup_{n \to \infty} \frac{|X \cap [1,n]|}{n}, $$ and $g$ the upper Banach (or uniform) density, viz. the function $$ \mathcal P(\mathbf N^+) \to \mathbf R: X \mapsto \lim_{n \to \infty} \max_{h \ge 0} \frac{|X \cap [1+h,n+h]|}{n}, $$ where the latter limit exists as a consequence of Fekete's lemma on subadditive real sequences. (Both of these functions are upper densities, as it was mentioned in the OP.)
Next, fix $a \in {]0,1]}$, and set $$ V_a := \bigcup_{n \ge 1} [\![a(2n-1)!+(1-a) (2n)!, (2n)!]\!] $$ and $$ X := V_a \cup (V_a^c \cap (2\cdot \mathbf N)) \quad\text{and}\quad Y := V_a \cup (V_a^c \cap (2 \cdot \mathbf N+1)), $$ where $V_a^c := \mathbf N^+ \setminus V_a$. It is clear that $X \cup Y = \mathbf N^+$ and $X \cap Y = V_a$. Moreover, we get from here that $$ f(X) \le f(V_a) + f(V_a^c \cap (2 \cdot \mathbf N)) \le a +\frac{1}{2}, $$ and similarly $f(Y) \le a + \frac{1}{2}$ and $f(V_a) = a$. On the other hand, $V_a$ contains arbitrarily large intervals of consecutive integers, hence $g(X) = g(Y) = g(V_a) = 1$. It follows that $$ 1+h(X \cap Y) = 1 + (\alpha +a^q(1-\alpha))^{\frac{1}{q}} $$ and $$ h(X) + h(Y) \le 2 \cdot (\alpha +(0.5+a)^q(1-\alpha))^{\frac{1}{q}}. $$ With this in hand, suppose to a contradiction that $1+h(A \cup B) \le h(A) + h(B)$ for all $A, B \subseteq \mathbf N^+$ such that $A \cup B = \mathbf N^+$, regardless of the actual values of the parameters $\alpha$ and $q$. Then, we have from the above that $$ 1 + \alpha^{\frac{1}{q}} \le 2 \cdot (\alpha +\left(0.5+a\right)^q(1-\alpha))^{\frac{1}{q}}. $$ This implies, in the limit as $a \to 0^+$, that we should have $$ 1 + \alpha^{\frac{1}{q}} \le (2^q\alpha + 1-\alpha)^{\frac{1}{q}} $$ for all $\alpha \in [0,1]$ and $q \in [1,\infty[$, which is, however, false (e.g., let $\alpha = \frac{1}{2}$ and $q = 2$).