Seems that this is possible. Here is a (non-constructive) proof. Suggestions are welcome.

The proof is inspired by Mazurkiewicz's argument. This is second version of the proof: it includes improvements in the set-theoretic argument suggested by Joel David Hamkins, and also hopefully clarifies some issues raised in comments. Thanks for the comments!

Goal: Construct a commutative group structure $\star$ on non-negative reals ${\mathbb R}^{\ge 0}$ such that $x\star y\le x+y$ and $x\star x=0$.

Remark: Note that $0$ is automatically a neutral element, and that such a commutative group is in fact a vector space over ${\mathbb F}_2$. Also, we automatically have the triangle inequality: $$x\star z=x\star y\star y\star z\le x\star y+y\star z.$$

Step 1: Let us order ${\mathbb R}^{\ge 0}$ in order type $c$ (continuum). Equivalently, we choose a bijection $\iota:[0,c)\to{\mathbb R}^{\ge 0}$, where $[0,c)$ is the set of ordinals smaller than $c$. Note that for any $\alpha < c$, we have $$|\iota([0,\alpha))| < c.$$

We may choose $\iota$ so that $\iota(0)=0$, although it is not strictly necessary.

Plan: For every $\alpha\le c$, we will construct a subset $S_\alpha\subset {\mathbb R}^{\ge 0}$ and a group operation $\star:S_\alpha\times S_\alpha\to S_\alpha$. The group operation will have the required properties: $S_\alpha$ is a vector space over $F_2$ with $0$ being the neutral element, and $x\star y\le x+y$. Besides it will also have the additional property that $S_\alpha$ is generated as a group by $\iota([0,\alpha))$ (in particular, the image is contained in $S_\alpha$). Moreover, if $\beta\prec\alpha$, $S_\beta$ is a subgroup of $S_\alpha$.

In particular, we get a group structure with required properties on $S_c={\mathbb R}^{\ge 0}$, as claimed.

Step 2: The construction proceeds by transfinite recursion. The base is $S_0=\lbrace 0\rbrace$ (generated by the empty set).

Step 3. Let us now define $S_\alpha$ assuming that $S_\beta$ is already defined for $\beta<\alpha$. If $\alpha$ is a limit ordinal, take $$S_\alpha=\bigcup_{\beta<\alpha}S_\beta.$$ Therefore, let us assume $\alpha=\beta+1$.

If $\iota(\alpha)\in S_\beta$, take $S_\alpha=S_\beta$.

Step 4. It remains to consider the case when $\alpha=\beta+1$ but $\iota(\alpha)\not\in S_\beta$. Since $I=\iota([0,\beta))$ generates $S_\beta$,
the cardinality of $S_\beta$ is at most the cardinality of the set of finite subsets of $I$. Therefore, $|S_\beta| < c$.

Fix a number $k$ between $0$ and $1$, to be chosen later. Define a function $f:{\mathbb R}^{\ge 0}\to{\mathbb R}^{\ge 0}$ by $$f(x)=\cases{\alpha+k f(x)=\cases{\iota(\alpha)+k x,&x\le\alpha\cr x+k\alpha&x\le\iota(\alpha)\cr x+k\iota(\alpha),&x>\alpha}.$$ &x>\iota(\alpha)}.$$Now choose k so that f(S_\beta)\cap S_\beta=\emptyset. This is possible because for every x,y\in S_\beta, the equation f(x)=y has at most one solution in k, so the set of prohibited values of k has cardinality at most |S_\beta\times S_\beta|. (We can use \iota to well-order the interval (0,1); we can then choose k to be the minimal acceptable value, so as to remove arbitrary choice.) Step 5. Now define S_\alpha=S_\beta\cup f(S_\beta) and set \iota(\alpha)\star x=f(x) for x\in S_\beta. The product naturally extends to all of S_\alpha:$$f(x)\star f(y)=x\star y\qquad f(x)\star y=y\star f(x)=f(x\star y).$$It is not hard to see that it has the required properties. First of all, S_\alpha is an isomorphic image of S_\beta\times({\mathbb Z}/2{\mathbb Z}); this takes care of group-theoretic requirement. It remains to check two inequalities: Step 5a:$$f(x)\star f(y)\le f(x)+f(y)\quad(x,y\in S_\beta),$$which is true because f(x)\ge x, so$$f(x)\star f(y)=x\star y\le x+y\le f(x)+f(y).$$Step 5b:$$f(x)\star y\le f(x)+y\quad(x,y\in S_\beta),$$which is true because f is increasing and f(x+t)\le f(x)+t, so$$f(x)\star y=f(x\star y)\le f(x+y)\le f(x)+y.$$That's it. 4 added 2 characters in body; added 8 characters in body; deleted 18 characters in body Seems that this is possible. Here is a (non-constructive) proof. Suggestions are welcome. The proof is inspired by Mazurkiewicz's argument. This is second version of the proof: it includes improvements in the set-theoretic argument suggested by Joel David Hamkins, and also hopefully clarifies some issues raised in comments. Thanks for the comments! Goal: Construct a commutative group structure \star on non-negative reals {\mathbb R}^{\ge 0} such that x\star y\le x+y and x\star x=0. Remark: Note that 0 is automatically a neutral element, and that such a commutative group is in fact a vector space over  {\mathbb F}_2{\mathbb F}_2 . Also, we automatically have the triangle inequality:$$x\star z=x\star y\star y\star z\le x\star y+y\star z.$$Step 1: Let us order {\mathbb R}^{\ge 0} in order type c (continuum). Equivalently, we choose a bijection \iota:[0,c)\to{\mathbb R}^{\ge 0}, where [0,c) is the set of ordinals smaller than c. Note that for any  \alpha < c , we have$$|\iota([0,\alpha))| < c.$$We may choose \iota so that \iota(0)=0, although it is not strictly necessary. Plan: For every \alpha\le c, we will construct a subset S_\alpha\subset {\mathbb R}^{\ge 0} and a group operation \star:S_\alpha\times S_\alpha\to S_\alpha. The group operation will have the required properties: S_\alpha is a vector space over {\mathbb F}2 F_2 with 0 being the neutral element, and x\star y\le x+y. Besides it will also have the additional property that S\alpha S_\alpha is generated as a group by \iota([0,\alpha)) (in particular, the image is contained in S_\alpha). Moreover, if \beta\prec\alpha, S_\beta is a subgroup of S_\alpha. In particular, we get a group structure with required properties on S_c={\mathbb R}^{\ge 0}, as claimed. Step 2: The construction proceeds by transfinite recursion. The base is S_0=\lbrace 0\rbrace (generated by the empty set). Step 3. Let us now define S_\alpha assuming that S_\beta is already defined for \beta<\alpha. If \alpha is a limit ordinal, take$$S_\alpha=\bigcup_{\beta<\alpha}S_\beta.$$Therefore, let us assume \alpha=\beta+1. If \iota(\alpha)\in S_\beta, take S_\alpha=S_\beta. Step 4. It remains to consider the case when \alpha=\beta+1 but \iota(\alpha)\not\in S_\beta. Since I=\iota([0,\beta)) generates S_\beta, the cardinality of S_\beta is at most the cardinality of the set of finite subsets of I. Therefore, |S_\beta| Fix a number k between 0 and 1, to be chosen later. Define a function f:{\mathbb R}^{\ge 0}\to{\mathbb R}^{\ge 0} by$$f(x)=\cases{\alpha+k x,&x\le\alpha\cr x+k\alpha,&x>\alpha}.$$Now choose k so that f(S_\beta)\cap S_\beta=\emptyset. This is possible because for every x,y\in S_\beta, the equation f(x)=y has at most one solution in k, so the set of prohibited values of k has cardinality at most |S_\beta\times S_\beta|. (We can use \iota to well-order the interval (0,1); we can then choose k to be the minimal acceptable value, so as to remove arbitrary choice.) Step 5. Now define S_\alpha=S_\beta\cup f(S_\beta) and set \iota(\alpha)\star x=f(x) for x\in S_\beta. The product naturally extends to all of S_\alpha:$$f(x)\star f(y)=x\star y\qquad f(x)\star y=y\star f(x)=f(x\star y).$$It is not hard to see that it has the required properties. First of all, S_\alpha is an isomorphic image of S_\beta\times({\mathbb Z}/2{\mathbb Z}); this takes care of group-theoretic requirement. It remains to check two inequalities: Step 5a:$$f(x)\star f(y)\le f(x)+f(y)\quad(x,y\in S_\beta),$$which is true because f(x)\ge x, so$$f(x)\star f(y)=x\star y\le x+y\le f(x)+f(y).$$Step 5b:$$f(x)\star y\le f(x)+y\quad(x,y\in S_\beta),$$which is true because f is increasing and f(x+t)\le f(x)+t, so$$f(x)\star y=f(x\star y)\le f(x+y)\le f(x)+y.$$That's it. 3 cleaned up set-theoretic argument Let me try a proof Seems that this is possible. It's going to be non-constructive Here is a (to put it mildly)non-constructive) proof. Improvements Suggestions are welcome. The proof is inspired by Mazurkiewicz's argument. This is second versionof the proof: it includes improvements inthe set-theoretic argument suggested by Joel David Hamkins, and also hopefully clarifies some issuesraised in comments. Thanks for the comments! (Edit: It Remark: Note that 0 is automatically satisfies a neutral element, and that such a commutative group is in facta vector space over {\mathbb F}_2. Also, we automatically have the$$x\star z=x\star y\star y\star z\le x\star y+y\star z.)$$z.$$

Step 1: Choose an Let us order $\succ$ on ${\mathbb R}^{\ge 0}$ such that in order type ${\mathbb R}^{\ge 0}$ is well-ordered and for any c$(continuum). Equivalently,we choose a bijection$x\in {\mathbb \iota:[0,c)\to{\mathbb R}^{\ge 0}$, where$[0,c)$is the set$\lbrace y:y\prec x\rbrace$has cardinality strictly less of ordinals smaller than continuum$c$. Note that for any$ \alpha < c $, we have$$|\iota([0,\alpha))| < c.$$ We may choose$\iota$so the ordinal type of that$\succ$\iota(0)=0$, although it is a cardinalnot strictly necessary.

Step 2: Consider the following triples

Plan: $(I,S,\star)$, where For every $S\subset{\mathbb R}^{\ge 0}$ is\alpha\le c$, we will construct a subset ,$\star:S\times S\to S$is S_\alpha\subset {\mathbb R}^{\ge 0}$ and a group operationon $S$ that satisfies our requirements, and\star:S_\alpha\times S_\alpha\to S_\alpha$. The group operation will have the required properties:$I\subset S$S_\alpha$ isa subset which is an initial interval for vector space over $\succ$ {\mathbb F}2$with$0$being the neutral element, and generates$S$x\star y\le x+y$. Besidesit will also have the additional property that $S\alpha$ is generated as a group .

The set of such triples by $\iota([0,\alpha))$(in particular, the image is naturally ordered. contained in $S_\alpha$). Moreover, any chain$(I_\alpha,S_\alpha,\star_\alpha)$ has an upper bound \beta\prec\alpha$, $$(\bigcup I_\alpha,\bigcup S_\alpha,\star),$$(a union of initial intervals S_\beta$ is an interval and a union subgroupof generating sets generates the union). By Zorn's Lemma$S_\alpha$.

In particular, there is we get a maximal element group structure with required properties on $(I,S,\star)$.S_c={\mathbb R}^{\ge 0}$, as claimed. Step 32: Let us show that The construction proceeds by transfinite recursion. The base is$S={\mathbb R}^{\ge 0}.$Assume S_0=\lbrace 0\rbrace$ (generated by the converseempty set).Note

Step 3. Let us now define $S_\alpha$ assuming that $I\subset {\mathbb R}^{\ge 0}$ S_\beta$is already defined for$\beta<\alpha$. If$\alpha$is a proper initial intervallimit ordinal, so its cardinality is strictly smaller than continuumtake$$S_\alpha=\bigcup_{\beta<\alpha}S_\beta.$$Therefore, let us assume$\alpha=\beta+1$. If$\iota(\alpha)\in S_\beta$, take$S_\alpha=S_\beta$. Step 4. It remains to consider the case when$\alpha=\beta+1$but$\iota(\alpha)\not\in S_\beta$.Since$I$I=\iota([0,\beta))$ generates $S$, S_\beta$, the cardinality of$S$(which S_\beta$ 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 bythe initial interval $\lbrace x:x\preceq\alpha\rbrace$I$. Therefore, so it would contradict maximality of$(I,S,\alpha)$.|S_\beta| Fix a number$c$k$ between $0$ and $1$, to be chosen later. Define a $$f(x)=\cases{\alpha+c f(x)=\cases{\alpha+k x,&x\le\alpha\cr x+c\alphax+k\alpha,&x>\alpha}.$$Now choose $c$ k$so that$f(S)\cap S=\emptyset$f(S_\beta)\cap S_\beta=\emptyset$. This is possible because for every $x,y\in S$S_\beta$, the equation$f(x)=y$has at most one solution in$c$, k$, so the set of prohibited values of $c$ k$has cardinality at most$|S\times S|$|S_\beta\times S_\beta|$. (We can use $\iota$ to well-order the interval $(0,1)$;we can then choose $k$ to be the minimal acceptable value, so as to remove arbitrary choice.)

Step 45. Now define $T=S\cup f(S)$ S_\alpha=S_\beta\cup f(S_\beta)$and set$\alpha\star \iota(\alpha)\star x=f(x)$for$x\in S$S_\beta$. The product naturally extends to all of $T$:

Indeed

First of all, we need $S_\alpha$ is an isomorphic image of $S_\beta\times({\mathbb Z}/2{\mathbb Z})$;this takes care of group-theoretic requirement. It remains to

Step 4a5a: $$f(x)\star f(y)\le f(x)+f(y)\quad(x,y\in S),$$

Step 4b5b: $$f(x)\star y\le f(x)+y\quad(x,y\in S),$$

That's it.

Edit: Forgot to mention that the proof is inspired by Mazurkiewicz's argument.