Let $T$ denote a commutative algebraic theory with a constant symbol. (We definitely need to assume that $T$ has a constant symbol, otherwise the algebraic theory of idempotent Abelian semigroups is an easy counterexample.)
Question. Is it a general principle that if an identity $\eta$ in the language of $T$ holds for the $T$-algebra freely generated by one element, then it is a theorem of $T$?
For example, this holds in the special case where $T$ is the equational theory of idempotent Abelian monoids. The $T$-algebra freely generated by the set $\{1\}$ is the Boolean domain $\mathbb{B} = \{0,1\}$. An identity in this language is a theorem iff each variable that occurs on the left also occurs on the right, and vice versa. For example, $x+(a+b)=(x+a)+(x+b)$ is a theorem, but $x+b=x$ is not. Let us now prove the claim for this special case; i.e. we will show that if $\mathbb{B} \models \eta$, then $T \vdash \eta$. We will use the following facts about $\mathbb{B}$:
$$1+1 = 0+1 = 1+0 = 1, \quad\quad 0+0 = 0$$
Proof. Suppose $\mathbb{B} \models \eta$ where $\eta$ is an identity in the set of variables $X$. Now suppose for a contradiction that there is an $x \in X$ that occurs on one side of $\eta$ but not the either. Since $\mathbb{B} \models \eta,$ hence $\mathbb{B}\models \eta/x$, where $\eta/x$ is obtained by replacing $x$ with $1$ and every variable except $x$ with $0$. But this using the above identities, we can show by induction that this makes one side of $\eta$ equal $0$ and the other side equal $1$. Hence $\mathbb{B} \models (0=1)$, a contradiction.
We deduce that if $\mathbb{B} \models \eta$ where $\eta$ is an identity in the set of variables $X$, then every $x \in X$ either occurs on both sides of $\eta$, or on neither side. Thus, $T \vdash \eta$.
I assume "commutative algebraic theory" means that the language should have one commutative binary operation, plus constants? Do you also require associativity?
Anyway, here's an associative counterexample: Let $L = \{\cdot, c\}$, and consider the theory axiomatized by $T$:
Now the free $T$-algebra on one generator has two elements, $\{a,c\}$, and multiplying any two elements gives $c$. Hence this algebra satisfies the identity $x\cdot y = c$.
But this identity doesn't follow from $T$. In particular, it doesn't hold of the free $T$-algebra on two generators.