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The Lorentz signature is just part of the theory; for example in a weak-field limit, we should reduce to special relativity, which is described using Lorentz signature (in order to talk about light, and also because Lorentz signature allows us to encode time and space into a single entity, and defines our notions of causality).

1. No, it's not in the Ricci tensor. Any smooth manifold admits a connection $\nabla$, and with respect to this connection we can consider its curvature tensor $R^a_{\,bcd}$, and from this by taking a contraction we can define the Ricci tensor $R_{ab}= R^s_{\,asb}$. And connections can be very arbitrary, not necessarily even arising from a pseudo-Riemannian metric.
2. I don't see any direct links.
3. No it's not in the stress energy tensor. For instance, suppose we have an arbitrary pseudo-Riemannian metric $g$ (arbitrary signature $(p,q)$). We can still consider its Einstein tensor $R_{ab}-\frac{1}{2}Rg_{ab}$. By setting this to $0$, we see that $\Bbb{R}^4$ with the standard flat pseudo-Riemannian metric $\eta_{p,q}$ is a solution. However, for this to bear any resemblance to (special) relativity we need a $(1,3)$ signature split.
4. See (3).
5. Yes.
6. You mentioned (5), so you're not looking at it wrong.

Also, when coming up with any particular solution, such as Schwarzschild, or Kerr or anything else, the ansatz already imposes Lorentz signature.

The Lorentz signature is just part of the theory; for example in a weak-field limit, we should reduce to special relativity, which is described using Lorentz signature (in order to talk about light, and also because Lorentz signature allows us to encode time and space into a single entity, and defines our notions of causality).

1. No, it's not in the Ricci tensor. Any smooth manifold admits a connection $\nabla$, and with respect to this connection we can consider its curvature tensor $R^a_{\,bcd}$, and from this by taking a contraction we can define the Ricci tensor $R_{ab}= R^s_{\,asb}$. And connections can be very arbitrary, not necessarily even arising from a pseudo-Riemannian metric.
2. I don't see any direct links.
3. No it's not in the stress energy tensor. For instance, suppose we have an arbitrary pseudo-Riemannian metric $g$ (arbitrary signature $(p,q)$). We can still consider its Einstein tensor $R_{ab}-\frac{1}{2}Rg_{ab}$. By setting this to $0$, we see that $\Bbb{R}^4$ with the standard flat pseudo-Riemannian metric $\eta_{p,q}$ is a solution. However, for this to bear any resemblance to (special) relativity we need a $(1,3)$ signature split.
4. See (3).
5. Yes (though I'd say it's an explicit assumption).
6. You mentioned (5), so you're not looking at it wrong.

Also, when coming up with any particular solution, such as Schwarzschild, or Kerr or anything else, the ansatz already imposes Lorentz signature.

The Lorentz signature is just part of the theory; for example in a weak-field limit, we should reduce to special relativity, which is described using Lorentz signature (in order to talk about light, and also because Lorentz signature allows us to encode time and space into a single entity, and defines our notions of causality).

1. No, it's not in the Ricci tensor. Any smooth manifold admits a connection $\nabla$, and with respect to this connection we can consider its curvature tensor $R^a_{\,bcd}$, and from this by taking a contraction we can define the Ricci tensor $R_{ab}= R^s_{\,asb}$. And connections can be very arbitrary, not necessarily even arising from a pseudo-Riemannian metric.
2. I don't see any direct links.
3. No it's not in the stress energy tensor. For instance, suppose we have an arbitrary Pseudo-RIemannian metric $g$ (arbitrary signature $(p,q)$). We can still consider its Einstein tensor $R_{ab}-\frac{1}{2}Rg_{ab}$. By setting this to $0$, we see that $\Bbb{R}^4$ with the standard flat pseudo-Riemannian metric $\eta_{p,q}$ is a solution. However, for this to bear any resemblance to (special) relativity we need a $(1,3)$ signature split.
4. See (3).
5. Yes.
6. You mentioned (5), so you're not looking at it wrong.

Also, when coming up with any particular solution, such as Schwarzschild, or Kerr or anything else, the ansatz already imposes Lorentz signature.

The Lorentz signature is just part of the theory; for example in a weak-field limit, we should reduce to special relativity, which is described using Lorentz signature (in order to talk about light, and also because Lorentz signature allows us to encode time and space into a single entity, and defines our notions of causality).

1. No, it's not in the Ricci tensor. Any smooth manifold admits a connection $\nabla$, and with respect to this connection we can consider its curvature tensor $R^a_{\,bcd}$, and from this by taking a contraction we can define the Ricci tensor $R_{ab}= R^s_{\,asb}$. And connections can be very arbitrary, not necessarily even arising from a pseudo-Riemannian metric.
2. I don't see any direct links.
3. No it's not in the stress energy tensor. For instance, suppose we have an arbitrary pseudo-Riemannian metric $g$ (arbitrary signature $(p,q)$). We can still consider its Einstein tensor $R_{ab}-\frac{1}{2}Rg_{ab}$. By setting this to $0$, we see that $\Bbb{R}^4$ with the standard flat pseudo-Riemannian metric $\eta_{p,q}$ is a solution. However, for this to bear any resemblance to (special) relativity we need a $(1,3)$ signature split.
4. See (3).
5. Yes.
6. You mentioned (5), so you're not looking at it wrong.

Also, when coming up with any particular solution, such as Schwarzschild, or Kerr or anything else, the ansatz already imposes Lorentz signature.

The Lorentz signature is just part of the theory; for example in a weak-field limit, we should reduce to special relativity, which is described using Lorentz signature (in order to talk about light, and also because Lorentz signature allows us to encode time and space into a single entity, and defines our notions of causality).

1. No, it's not in the Ricci tensor. Any smooth manifold admits a connection $\nabla$, and with respect to this connection we can consider its curvature tensor $R^a_{\,bcd}$, and from this by taking a contraction we can define the Ricci tensor $R_{ab}= R^s_{\,asb}$. And connections can be very arbitrary, not necessarily even arising from a pseudo-Riemannian metric.
2. I don't see any direct links.
3. No it's not in the stress energy tensor. For instance, suppose we have an arbitrary Pseudo-RIemannian metric $g$ (arbitrary signature $(p,q)$). We can still consider its Einstein tensor $R_{ab}-\frac{1}{2}Rg_{ab}$. By setting this to $0$, we see that $\Bbb{R}^4$ with the standard flat pseudo-Riemannian metric $\eta_{p,q}$ is a solution. However, for this to bear any resemblance to (special) relativity we need a $(1,3)$ signature split.
4. See (3).
5. Yes.
6. You mentioned (5), so you're not looking at it wrong.

Also, when coming up with any particular solution, such as Schwarzschild, or Kerr or anything else, the anszatz already imposes Lorentz signature.

The Lorentz signature is just part of the theory; for example in a weak-field limit, we should reduce to special relativity, which is described using Lorentz signature (in order to talk about light, and also because Lorentz signature allows us to encode time and space into a single entity, and defines our notions of causality).

1. No, it's not in the Ricci tensor. Any smooth manifold admits a connection $\nabla$, and with respect to this connection we can consider its curvature tensor $R^a_{\,bcd}$, and from this by taking a contraction we can define the Ricci tensor $R_{ab}= R^s_{\,asb}$. And connections can be very arbitrary, not necessarily even arising from a pseudo-Riemannian metric.
2. I don't see any direct links.
3. No it's not in the stress energy tensor. For instance, suppose we have an arbitrary Pseudo-RIemannian metric $g$ (arbitrary signature $(p,q)$). We can still consider its Einstein tensor $R_{ab}-\frac{1}{2}Rg_{ab}$. By setting this to $0$, we see that $\Bbb{R}^4$ with the standard flat pseudo-Riemannian metric $\eta_{p,q}$ is a solution. However, for this to bear any resemblance to (special) relativity we need a $(1,3)$ signature split.
4. See (3).
5. Yes.
6. You mentioned (5), so you're not looking at it wrong.

Also, when coming up with any particular solution, such as Schwarzschild, or Kerr or anything else, the ansatz already imposes Lorentz signature.