Suppose we have
$$\varphi(x) = \frac{e^x}{1+x}, \ \ \ x>0.$$
Does there exist a constant $C>0$ and a positive function $v(x)$ such that $\varphi(x+y) \leq Cv(x)\varphi(y)$ for all $x,y \in \mathbb{R}^+$ with $v(x+y) \leq v(x)v(y)$ for all $x,y \in \mathbb{R}$ and $ve^{-|\cdot|} \in L^p(\mathbb{R})$ for $p>0$?
So far all we know is that
$$\varphi(x+y) = \frac{e^xe^y}{1+x+y} \leq e^x\varphi(y)$$
which takes care of the first two properties if we take $v(x) = e^x$, but $ve^{-|\cdot|} \notin L^p(\mathbb{R})$.
From $\varphi(x+y) \le C\,v(x)\,\varphi(y)$ we get $$ v(x)\,e^{-x}\ge\frac1C\,\frac{\varphi(x+y)}{\varphi(y)}\,e^{-x}=\frac1C\,\frac{1+y}{1+x+y},\quad x,y>0. $$ Letting $y\to\infty$, we see that $v(x)\,e^{-x}\ge1/C$, which is not in $L^p$.