Problem
Show that a product metric $\big(M_1\times M_2,g_1+g_2\big)$ satisfies $\text{Ric}^0=0$, (where $\text{Ric}^0$ is the traceless Ricci tensor) if and only if both original metrics are Einstein with $\frac{s_1}{\dim M_1}=\frac{s_2}{\dim M_2}$.
Thoughts
This is what I have so far, I can't tell I'm doing everything right though, but to me it at least seems right
Starting with the converse.
$(\Leftarrow)$ Let $\nabla_1,\nabla_2$ be the Levi-Civita connection forms for $M_1,M_2$, respectively. Let $X\in\Gamma(T(M_1\times M_2))$ and $\pi_i:T(M_1\times M_2)\to TM_i$ be the canonical projections. Define $$ \nabla X:=\nabla_1\pi_1(X)+\nabla_2\pi_2(X). $$ Note that $\pi_1,\pi_2$ are surjective, so $\nabla$ is well defined. Since $\nabla_1,\nabla_2$ are torsion-free and are compatible with the metrics $g_1,g_2$, respectively, it is clear that $\nabla$ also has these properties with respect to the product metric $g:=g_1+g_2$; thus the fundamental lemma of Riemannian geometry asserts that $\nabla$ is the Levi-Civita form on $M_1\times M_2$. Further, by linearity of $\nabla_1,\nabla_2$ in both factors and noting that $[\pi_i(V),\pi_j(W)]=0$ when $i\neq j$ for any two $V,W\in\Gamma(T(M_1\times M_2))$ (from here on, let $X_i,Y_i,Z_i$ denote that projection of $X,Y,Z$, respectively, onto the $i$th component), \begin{align*} \Theta(X,Y)Z&=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z\\ &=\nabla_X\big({\nabla_1}\big)_{Y_1}Z_1+\nabla_X\big(\nabla_2\big)_{Y_2}Z_2-\big(\nabla_1\big)_{[X_1,Y_1]}Z_1-\big(\nabla_2\big)_{[X_2,Y_2]}Z_2\\ &=\big(\nabla_1\big)_{X_1}\big(\nabla_1\big)_{Y_1}Z_1-\big(\nabla_1\big)_{Y_1}\big(\nabla_1\big)_{X_1}Z_1-\big(\nabla_1\big)_{[X_1,Y_1]}Z_1\\ &\hspace{1 in}+\big(\nabla_2\big)_{X_2}\big(\nabla_2\big)_{Y_2}Z_2-\big(\nabla_2\big)_{Y_2}\big(\nabla_2\big)_{X_2}Z_2-\big(\nabla_2\big)_{[X_2,Y_2]}Z_2\\ &=\Theta_1(X_1,Y_1)Z_1+\Theta_2(X_2,Y_2)Z_2. \end{align*} From this it is clear that $$ R(X,Y,Z,W)=R'(X_1,Y_1,Z_1,W_1)+R''(X_2,Y_2,Z_2,W_2), $$ where $R',R''$ belong to $M_1,M_2$, respectively (for lack of better notation); in coordinates this is $$ R_{ijkl}=R'_{ijkl}+R''_{ijkl}. $$ Thus (again, I apologize for the terrible notation) letting $\dim M_1=m_1$ and $\dim M_2=m_2$, \begin{align*} \text{Ric}^0&=\text{Ric}'+\text{Ric}''-\frac{s_1+s_2}{m_1+m_2}(g_1+g_2)\\ &=\text{Ric}'+\text{Ric}''-\frac{s_1g_1+s_1g_2+s_2g_1+s_2g_2}{m_1+m_2}\\ &=\frac{(m_1+m_2)\text{Ric}'+(m_1+m_2)\text{Ric}''-m_1\text{Ric}'-m_1\text{Ric}''-m_2\text{Ric}'-m_2\text{Ric}''}{m_1+m_2}\\ &=0. \end{align*}
Questions
Edit: I've updated my approach. I now have a new request if anyone were to be so kind: is this logic correct? Am I making any egregious assumptions? Also, in the $(\Rightarrow)$ direction, is it wrong to define $\nabla_1=(\nabla\iota_1X)$ and $\nabla_2=(\nabla\iota_2 Y)$ for $X\in\Gamma(M_1),Y\in\Gamma(M_2)$ and $\iota_1\iota_2$ the inclusion maps? I believe so, but I'm not very confident. Again, any help is greatly appreciated.