I know that there are some questions regarding symmetric operads, but I feel like asking yet another question.
Consider the operad $A$ with $A(n)=k[S_n]$. Let us denote the simple transpositions in $S_n$ by $s_1^{(n)},\dotsc,s_{n-1}^{(n)}$. If I have understood things correctly, algebras over this operad should be associative $k$-algebras.
Now, let's verify that there is a single multiplication indeed. Let $\mu = 1e \in A(2)$ be a fixed binary multiplication ($e$ is the empty word) and $\mu s$ be the multiplication with reversed arguments.
Now, I want to show that $\mu(\mu\otimes 1)$ and $\mu(1\otimes\mu)$ are the same. Let $M$ be an $A$-algebra and $x, y, z\in M$. Let $\theta_1=\mu$ and $\theta_2=1$.
Then
$$\begin{aligned} &\phantom{{}={}}\mu(1\otimes \mu) (x\otimes y\otimes z)\\ &= \mu(\theta_2\otimes \theta_1) (x\otimes y\otimes z)\\ &= \mu(\theta_{s^{(2)}(1)} \theta_{s^{(2)}(2)}) (x\otimes y\otimes z)\\ &= \mu s (\theta_1, \theta_2) s^{(3)}_1 s^{(3)}_2 (x\otimes y\otimes z)\\ &= \mu s (\mu\otimes 1)(z\otimes y\otimes x)\end{aligned}$$
which doesn't help at all.
So, how do I show that multiplication in an $A$-algebra needs to be associative?
Edit: To make it a little bit more clear: From the definition, I know that $S_3$ acts freely on $A(3)$. Thus, I know that $\mu(\mu\otimes 1)$ can be written as a linear combination
$$\mu(\mu\otimes 1) = \sum_{\sigma\in S_n} \lambda_\sigma \mu(1\otimes\mu)\sigma$$
for $\lambda_\sigma\in k$. If I am working with set-valued operads, I even have $\mu(\mu, 1) = \mu(1,\mu)\sigma$ for some $\sigma\in S_3$. However, how do I know that actually $\mu(\mu\otimes 1) = \mu(1\otimes \mu)$ from that?