Playing around with EFE: Ricci scalar into stress-energy density? Moving $\Lambda$ into the energy density:

Einstein's equations with a cosmological constant $\Lambda$ (setting $c=1$), as are used for example in the solar system, are
\begin{equation}
G^\mu_{\ \ \nu} + \Lambda \delta^\mu_{\ \ \nu}= 8 \pi G_N T^\mu_{\ \ \nu}
\end{equation}
We can move the cosmological constant term to the right hand side as follows
\begin{equation}
G^\mu_{\ \ \nu} = 8 \pi G_N \left( T^\mu_{\ \ \nu} + [T_\Lambda]^\mu_{\ \ \nu} \right)
\end{equation}
where we have defined an effective stress-energy tensor
\begin{equation}
[T_\Lambda]^\mu_{\ \ \nu} \equiv - \frac{\Lambda}{8 \pi G_N} \delta^\mu_{\ \ \nu}
\end{equation}
The definition of energy density is $\rho = - T^0_{\ \ 0} = \frac{\Lambda}{8\pi G_N}$, and pressure is $p = T^1_{1} = T^2_2 = T^3_3 = -\frac{\Lambda}{8\pi G_N}$. From this you can see that by definition, a cosmological constant has $\rho = - p$.
This is from the answer to another question, I will link it in the comments.
Moving $R$ into the energy density?

I'm wondering whether it is possible and allowed and makes sense to do the same mathematical operations with the Ricci scalar and whether one can gain some insight from that?
Einstein's equations without a cosmological constant $\Lambda$ (setting $c=1$) are
\begin{equation}
R_{\mu\nu} -\frac{1}{2} g_{\mu\nu}R= \kappa T_{\mu\nu}
\end{equation}
Moving the $R$-term to the right hand side results in
\begin{equation}
R_{\mu\nu} = \kappa\left( T_{\mu\nu} - [T_R]_{\mu\nu} \right)
\end{equation}
where we have defined an effective stress-energy tensor
\begin{equation}
[T_R]_{\mu\nu} \equiv - \frac{R}{2\kappa} \delta^\mu_{\ \ \nu}
\end{equation}
The definition of energy density is $\rho = - T^0_{\ \ 0} = -\frac{R}{2\kappa}$, and pressure is $p = T^1_{1} = T^2_2 = T^3_3 = \frac{R}{2\kappa}$.
This results in a pressure - can this be interpreted as the gravitational force (direction to the centre of gravity)? Or is this nonsense, then why is it nonsense?
 A: In principle, there is no gain from doing this procedure. Notice that when writing
$$G_{\mu\nu} = 8\pi T_{\mu\nu}$$
we have the huge advantage that the LHS is purely geometric, while the RHS is purely model-dependent matter. Each of the sides of the equation have completely different meanings. Nothing in the LHS is experimental (well, perhaps the dimension and signature of spacetime are), while everything on the RHS is experimental.
However, there is a trick that does something similar to what you want, but in a slightly more complicated way to get to a more interesting result. Let us begin with the field equations and take their trace. We have
\begin{align}
R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} &= 8\pi T_{\mu\nu}, \\
R_{\mu\nu}g^{\mu\nu} - \frac{1}{2} R g_{\mu\nu}g^{\mu\nu} &= 8\pi T_{\mu\nu}g^{\mu\nu}, \\
R - 2 R &= 8\pi T, \\
R &= -8\pi T,
\end{align}
where I defined $T = T_{\mu\nu}g^{\mu\nu}$ and used the fact that, in $d$-dimensions, $g_{\mu\nu}g^{\mu\nu} = d$.
Notice now that using this result we can write
\begin{align}
R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} &= 8\pi T_{\mu\nu}, \\
R_{\mu\nu} + 4\pi T g_{\mu\nu} &= 8\pi T_{\mu\nu}, \\
R_{\mu\nu} &= 8\pi \left(T_{\mu\nu} - \frac{1}{2} T g_{\mu\nu}\right),
\end{align}
which resembles the result you presented, but it is still different. Notice that, just like the original form of the Einstein Field Equations, the LHS is purely geometric, the RHS is model-dependent.
This equation can be quite useful in some situations where the stress-energy tensor is known to be traceless. For example, when all matter under consideration is electromagnetic. In this situation, we can immediately remove the trace term and get to
$$R_{\mu\nu} = 8\pi T_{\mu\nu}, \tag{$T_{\mu\nu}$ is trace-free}$$
providing a simpler version of the equations. This can be used, for example, when solving for the Reissner–Nordström metric. Instead of computing the Ricci scalar abstractly to solve the Einstein Equations, one can simply drop it by knowing that the stress tensor is traceless.
