Let me try a proof. It's going to be non-constructive (to put it mildly). Improvements are welcome.
Goal: Construct a commutative group structure $\star$ on non-negative reals ${\mathbb R}^{\ge 0}$ such that $x\star y\le x+y$ and $x\star x=0$.
Step 1: Choose a complete order $\succ$ on ${\mathbb R}^{\ge 0}$ such that for any $x\in {\mathbb R}^{\ge 0}$, the set $\lbrace y:y\prec x\rbrace$ has cardinality strictly less than continuum, so the ordinal type of $\succ$ is a cardinal.
Step 2: Consider the following triples: $(I,S,\star)$, where $S\subset{\mathbb R}^{\ge 0}$ is a subset, $\star:S\times S\to S$ is a group operation on $S$ that satisfies our requirements, and $I\subset S$ is a subset which is an initial interval for $\succ$ and generates $S$ as a group.
The set of such triples is naturally ordered. Moreover, any chain $(I_\alpha,S_\alpha,\star_\alpha)$ has an upper bound $$(\bigcup I_\alpha,\bigcup S_\alpha,\star),$$ (a union of initial intervals is an interval and a union of generating sets generates the union). By Zorn's Lemma, there is a maximal element $(I,S,\star)$.
Step 3: Let us show that $S={\mathbb R}^{\ge 0}.$ Assume the converse. Note that $I\subset {\mathbb R}^{\ge 0}$ is a proper initial interval, so its cardinality is strictly smaller than continuum. Since $I$ generates $S$, the cardinality of $S$ (which is at most the cardinality of the set of finite subsets of $I$) is also strictly less than continuum.
Let $\alpha$ be the smallest element of ${\mathbb R}^{\ge 0}-S$. We will construct a set $T\supset S\cup\lbrace\alpha\rbrace$ and an extension of $\star$ to a group operation (with required properties) on $T$ such that $T=S\cup S\star\alpha$. Then $T$ is generated by the initial interval $\lbrace x:x\preceq\alpha\rbrace$, so it would contradict maximality of $(I,S,\alpha)$.
Fix a number $c$ between $0$ and $1$, to be chosen later. Define a function $f:{\mathbb R}^{\ge 0}\to{\mathbb R}^{\ge 0}$ by $$f(x)=\cases{\alpha+c x,&x\le\alpha\cr x+c\alpha,&x>\alpha}.$$ Now choose $c$ so that $f(S)\cap S=\emptyset$. This is possible because for every $x,y\in S$, the equation $f(x)=y$ has at most one solution in $c$, so the set of prohibited values of $c$ has cardinality at most $|S\times S|$.
Step 4. Now define $T=S\cup f(S)$ and set $\alpha\star x=f(x)$ for $x\in S$. The product naturally extends to all of $T$: $$f(x)\star f(y)=x\star y\qquad f(x)\star y=y\star f(x)=f(x\star y).$$ It is not hard to see that it has the required properties.
Indeed, we need to check two inequalities:
Step 4a: $$f(x)\star f(y)\le f(x)+f(y)\quad(x,y\in S),$$ which is true because $f(x)\ge x$, so $$f(x)\star f(y)=x\star y\le x+y\le f(x)+f(y).$$
Step 4b: $$f(x)\star y\le f(x)+y\quad(x,y\in S),$$ which is true because $f$ is increasing and $f(x+t)\le f(x)+t$, so $$f(x)\star y=f(x\star y)\le f(x+y)\le f(x)+y.$$
That's it.