I am trying to transform the ODE $$\frac{1}{\sin(\theta)}\frac{d}{d\theta}\left(\sin(\theta)\frac{dS}{d\theta}\right)+\lambda S=0,$$ in to Legendre's equation $$(1-\mu^2)\frac{d^2S}{d\mu^2}-2\mu\frac{dS}{d\mu}+n(n+1)S=0$$ when $\lambda=n(n+1)$ for $n=0,1,2..$ and $\mu=\cos(\theta)$.
I calculated that, \begin{align} \frac{dS}{d\theta}&=\frac{dS}{d\mu}\frac{d\mu}{d\theta}=-\sin(\theta)\frac{dS}{d\mu} \\ \frac{d^2S}{d\theta^2}&=\frac{d}{d\theta}\left(\frac{dS}{d\theta}\right)=-\cos(\theta)\frac{d^2S}{d\mu^2}. \end{align} Then, \begin{align} \frac{1}{\sin(\theta)}\left(\sin(\theta)\frac{d^2S}{d\theta^2}+\cos(\theta)\frac{dS}{d\theta}\right)+S\lambda&=0 \\ -\mu\frac{d^2S}{d\mu^2}-\mu\frac{dS}{d\mu}+n(n+1)S&=0. \end{align} But at this point, I don't see how Legendre's equation is possible. I know that the double derivative is calculated incorrectly, but I don't know how to rectify this.
You made a mistake in deriving an expression for $\frac{d^2S}{d\theta^2}$. Let's have a look: $$\frac{d^2S}{d\theta^2}=\frac{d}{d\theta}\left(\frac{dS}{d\theta}\right)=\frac{d}{d\theta}\left(-sin(\theta)\frac{dS}{d\mu}\right)=-\cos(\theta)\frac{dS}{d\mu} - sin(\theta)\frac{d}{d\theta}\left(\frac{dS}{d\mu}\right)=\\-\mu\frac{dS}{d\mu}- sin(\theta)\frac{d}{d\mu}\left(\frac{dS}{d\mu}\right)\cdot\frac{d\mu}{d\theta} =-\mu\frac{dS}{d\mu} + sin^2(\theta)\frac{d^2S}{d\mu^2}=\\-\mu\frac{dS}{d\mu} + (1-\mu^2)\frac{d^2S}{d\mu^2}$$ Plugging this into the original ODE you will get the Legendre differential equation.