The most powerful behaviour of a tensor equation is precisely this feature that if the tensor components of a tensor equation holds in one frame then in another frame, up to a coordinate transformation this equation is also valid.
The simplest way to see that is the following:
Consider the tensor equation:
$$A^{\mu \nu} - B^{\mu \nu} = C^{\mu \nu} \tag{1} $$
If the components of $$A^{\mu \nu} - B^{\mu \nu} = 0$$ in the frame $S$, then in a frame $S'$ this components are also zero.
Proof:
$$A'^{\mu \nu} - B'^{\mu \nu} = \frac{\partial x'^{\mu}}{\partial x^{\delta}}\frac{\partial x'^{\nu}}{\partial x^{\gamma}}A^{\delta \gamma} - \frac{\partial x'^{\mu}}{\partial x^{\delta}}\frac{\partial x'^{\nu}}{\partial x^{\gamma}}B^{\delta \gamma} = \frac{\partial x'^{\mu}}{\partial x^{\delta}}\frac{\partial x'^{\nu}}{\partial x^{\gamma}}(A^{\delta \gamma} - B^{\delta \gamma}) = \frac{\partial x'^{\mu}}{\partial x^{\delta}}\frac{\partial x'^{\nu}}{\partial x^{\gamma}}(0) \implies $$
$$\implies A'^{\mu \nu} - B'^{\mu \nu} = 0 $$
In the case of Bianchi Identities with a ($LIF = S$) coordinate system (which the Christoffell symbols can be made zero, but not their derivatives) you just need to perform another general coordinate transformation for Riemann tensor and covariant derivatives to see that the equation is valid in the $S' \neq (LIF)$