Consider the free group $F$ generated by $a,b$ with the new operation: $u*v=$ the shortlex smallest word among $uv$ and $vu$ assumming $a>b>a^{-1}>b^{-1}$. This is a commutative magma. Consider the decomposition $F=F_a\cup F_b\cup F_{a^{-1}}\cup F_{b^{-1}}\cup \{1\}$ where $F_x$ is the set of (reduced) words starting with $x$. Would not this be a paradoxical decomposition of the magma? This would show that the magma does not have an invariant mean.
Update Here is the proof. Suppose that there exists an invariant mean $m$. Note that for every reduced word $au$, $a^{-1}*au=u$ because $a$ is the biggest letter. Hence $a^{-1}F_a=F_b\cup F_{b^{-1}}\cup F_a\cup \{1\}$. Therefore $m(F_b)=m(F_{b^{-1}})=m(\{1\})=0$. Consider the set $F_{b,b'}$ of reduced words starting with $b$ and ending not in $b$. Then $b^{-1}F_{b,b'}$ contains all words starting in $a$ or $a^{-1}$ and ending not in $b$. Therefore the set of all words starting with $a$ or $a^{-1}$ and ending in $b$ has a full measure. Taking a word starting in $b^{-1}$ and ending in $b$ and multiplying it with $b$ on the left produces any word ending in $b$. Therefore the set of such words has measure 0, a contradiction.