I came to your question after looking for the same proof and not finding it anywhere. Following StijnDeVuyst suggestion, I managed to prove it using integration by parts. I will omit the integration limits, just keep in mind they exist.

$$\int \int uv \ dC(u, v) = \int\int uv C'(u,v) \ dudv,$$ where $C'(u,v) = \frac{\partial^2}{\partial u \partial v}C(u,v)$.

\begin{equation} \tag{1} \label{main} \int\int uv C'(u,v) \ dudv = \int v \left(\int u C'(u,v) du \right) dv. \end{equation}

Solving the inner integral \begin{align*} \int_0^1 u C'(u,v) du &= u\frac{\partial}{\partial v}C(u,v) \big\rvert_{u=0}^1 - \int_0^1 \frac{\partial}{\partial v} C(u,v) \ du\\ &= 1-\int_0^1 \frac{\partial}{\partial v} C(u,v) \ du\\ \end{align*}

where the first equality is integration by parts on $u$ and the last equality comes from the fact that $C(1, v) = v$.

Substituting in \ref{main} we have \begin{align*} \int \int uvC'(u,v) \ dudv &= \int v \left(1-\int \frac{\partial}{\partial v} C(u,v) \ du \right)dv\\ &= \int_0^1 v \ dv - \int\int v \frac{\partial}{\partial v}C(u,v) \ dudv\\ &= \frac{1}{2}-\int\left(\int v \frac{\partial}{\partial v} C(u,v) \ dv \right) du. \tag{2}\label{segunda} \end{align*}

Solving the inner integral with integration by parts on $v$ we get $$ \int_0^1 v \frac{\partial}{\partial v} C(u,v) \ dv = u - \int_0^1C(u,v) \ dv$$

and substituting in \ref{segunda} we have (now including the integration limits) \begin{align*} \int_0^1\int_0^1 uv \ dC(u,v) &= \int_0^1 \int_0^1 uvC'(u,v) \ dudv\\ &= \frac{1}{2} - \int_0^1 u \ du + \int_0^1 \int_0^1 C(u,v) \ dv du\\ &= \int_0^1 \int_0^1 C(u,v) \ du dv \end{align*}

which proves the identity \begin{align*} \rho_S &= 12\int_0^1 \int_0^1 uv \ dC(u,v) - 3\\ &= 12\int_0^1\int_0^1 C(u,v) \ dudv - 3. \end{align*}

I came to your question after looking for the same proof and not finding it anywhere. Following StijnDeVuyst suggestion, I managed to prove it using integration by parts. I will omit the integration limits, just keep in mind they exist.

$$\int \int uv \ dC(u, v) = \int\int uv C'(u,v) \ dudv,$$ where $C'(u,v) = \frac{\partial^2}{\partial u \partial v}C(u,v)$.

\begin{equation} \tag{1} \label{main} \int\int uv C'(u,v) \ dudv = \int v \left(\int u C'(u,v) du \right) dv. \end{equation}

Solving the inner integral \begin{align*} \int_0^1 u C'(u,v) du &= u\frac{\partial}{\partial v}C(u,v) \big\rvert_{u=0}^1 - \int_0^1 \frac{\partial}{\partial v} C(u,v) \ du\\ &= 1-\int_0^1 \frac{\partial}{\partial v} C(u,v) \ du\\ \end{align*}

where the first equality is integration by parts on $u$ and the last equality comes from the fact that $C(1, v) = v$.

Substituting in \ref{main} we have \begin{align*} \int \int uvC'(u,v) \ dudv &= \int v \left(1-\int \frac{\partial}{\partial v} C(u,v) \ du \right)dv\\ &= \int_0^1 v \ dv - \int\int v \frac{\partial}{\partial v}C(u,v) \ dudv\\ &= \frac{1}{2}-\int\left(\int v \frac{\partial}{\partial v} C(u,v) \ dv \right) du. \tag{2}\label{segunda} \end{align*}

Solving the inner integral with integration by parts on $v$ we get $$ \int_0^1 v \frac{\partial}{\partial v} C(u,v) \ dv = u - \int_0^1C(u,v) \ dv$$

and substituting in \ref{segunda} we have (now including the integration limits) \begin{align*} \int_0^1\int_0^1 uv \ dC(u,v) &= \int_0^1 \int_0^1 uvC'(u,v) \ dudv\\ &= \frac{1}{2} - \int_0^1 u \ du + \int_0^1 \int_0^1 C(u,v) \ dv du\\ &= \int_0^1 \int_0^1 C(u,v) \ du dv \end{align*}

which proves the identity \begin{align*} \rho_S &= 12\int_0^1 \int_0^1 uv \ dC(u,v) - 3\\ &= 12\int_0^1\int_0^1 C(u,v) \ dudv - 3. \end{align*}

This proof assumes that the copula $C$ has density given by $\frac{\partial^2}{\partial u\partial v}C(u,v)$. When this is not the case, check Theorem 5.1.1 and Corollary 5.1.2 from An Introduction to Copulas by Roger B. Nelsen.