Assume $R$ is a unital commutative ring, and assume that $M$ is a cyclic $R$-module.
By definition, the symmetric algebra on $M$, we define as $$\mathcal{S}(M) := \mathcal{T}(M)/\mathcal{C}(M).$$
Now, we define $$\mathcal{C}(M) = \langle m_1 \otimes \cdots \otimes m_n - m_{\sigma(1)} \otimes \cdots \otimes m_{\sigma(n)}\;|\; m_i \in M,\sigma \in S_n \rangle.$$
Now since $$\mathcal{C}(M) \subset \mathcal{T}(M)$$ all elements in $$\mathcal{C}(M)$$ will be a finite direct sum of such elements.
Now, we note that $$M = \langle m \rangle = Rm = \{rm \;|\; r \in R\}$$ for some $m \in M$.
Hence for an arbitrary element $s \in \mathcal{C}(M)$, we see that the generators for $s$ on the form $$ m_1 \otimes \cdots \otimes m_n - m_{\sigma(1)} \otimes \cdots \otimes m_{\sigma(n)}$$ $$= r_1m \otimes \cdots \otimes r_nm - r_{\sigma(1)}m \otimes \cdots \otimes r_{\sigma(n)}m$$ $$\underbrace{=}_{\text{properties of tensor-product}} (r_1\cdots r_n)(m \otimes \cdots \otimes m)-(r_{\sigma(1)} \cdots r_{\sigma(n)}(m \otimes \cdots \otimes m) = 0$$
where I in the last step use that $R$ is commutative so that $$r_1 \cdots r_n = r_{\sigma(1)} \cdots r_{\sigma(n)}$$ and it follows that $s = 0$, and since $s$ was arbitrary, we see that $$\mathcal{C}(M) = \{0\}$$ so that $$\mathcal{S}(M) = \mathcal{T}(M)/\{0\} \cong \mathcal{T}(M).$$
Is this proof correct? If not, what is wrong?