Are all additionlike operations on $\mathbb{R}$ of this form?

Alexis Olson showed in his answer that the maps $x \star y = a x + b y$ with $a,b \geq 0$ satisfy axioms $0$ and $1$. I will show that these are the only ones.

Denote by $m$ the map $m : \mathbb{R}^2 \to \mathbb{R}$ given by $m(x,y)= x\star y$. Then Axiom $0$ is actually equivalent to the additivity of $m$.

Indeed, writing $z=(x,y)$ and $z' = (x', y')$, Axiom $0$ gives $m(z-z') = m(z)-m(z')$ for all $z, z'$. In particular, $m(0) = m(0 - 0 ) = m(0) - m(0)=0$. This in turn gives $m(- z) = m(0 -z) = m(0) - m(z) = - m(z)$ for all $z$. Thus we conclude that for any $z, z'$ we have $m(z+z') = m(z - (-z')) = m(z) - m(-z')) = m(z) + m(z')$. Conversely, additivity clearly implies Axiom $0$.

Since $m$ is additive, we have $m(x,y) = m(x,0) + m(0,y)$ for all $x,y$. Thus $m$ can be written as a sum $m(x,y) = f(x) + g(y)$, where $f,g: \mathbb{R} \to \mathbb{R}$ are both additive.

To prove the claim, we must show that $f$ and $g$ are $\mathbb{R}$-linear. Assuming the axiom of choice, there exist additive maps on $\mathbb{R}$ that are not $\mathbb{R}$-linear, but they are highly pathological (see here for a review of additive maps on $\mathbb{R}$). I claim that Axiom $1$ prevents this from happening.

Indeed, an additive map on $\mathbb{R}$ that is not $\mathbb{R}$-linear has the property that its graph is dense in $\mathbb{R}^2$ (a proof is given on the Wikipedia page). However, an immediate consequence of Axiom $1$ is that if $x \geq 0$ and $y \geq 0$ then $| x \star y | \leq x \star y$, and in particular $x \star y \geq 0$. Thus $f$ and $g$ both map $[0,\infty)$ to itself. Therefore, their graphs cannot be dense in $\mathbb{R}^2$ and so they must be linear. We conclude that $m$ is given by $m(x,y) = a x + b y$ for some $a,b \in \mathbb{R}$, and since $f$ and $g$ map $[0,\infty)$ to itself we must have $a,b \geq 0$.

Any linear map with non-negative coefficients will satisfy these axioms.

If $x \star y = ax + by$ with $a,b \ge 0$, then

\begin{eqnarray} (x\star y) - (x' \star y') &=& (ax + by) - (ax' - by')\\ &=& a(x-x') - b(y-y')\\ &=& (x-x') \star (y-y') \end{eqnarray}

and

$$|x \star y| = |ax + by| \le |ax| + |by| = a|x| + b|y| = |x|\star|y|.$$

Your characterization is the special case where $a = b = k$.