Yep. I'll explain that positive pressure x positive volume = positive energy
Dimensional analysis doesn't work that way. Just because energy has the same units as pressure times volume doesn't mean that energy
is pressure times volume.
and give you an explanation as to why the stress-energy density gradient causes gravity.
Your explanation would necessarily be wrong (or at the very least, flat out contradict GR) since it is the stress-energy tensor, and not any of its gradients, that appears explicitly in the Einstein field equation.
The Einstein field equation (without the cosmological constant) is normally expressed as
$$R_{\mu\nu} \,-\, \frac{1}{2} R g_{\mu\nu} \,=\, \kappa T_{\mu\nu}$$
where $$R_{\mu\nu}$$ is the Ricci tensor, $$R \,=\, g^{\mu\nu} R_{\mu\nu}$$ is the Ricci scalar, and $$T_{\mu\nu}$$ is the stress-energy tensor, as you should already know. It is well known, and easy to show, that the EFE can alternatively be written as
$$R_{\mu\nu} \,=\, \kappa \bigl( T_{\mu\nu} \,-\, \frac{1}{2} T g_{\mu\nu} \bigr)$$
where $$T \,=\, g^{\mu\nu} T_{\mu\nu}$$ is the trace of the stress-energy tensor. This tells you what sort of stress-energy density you need in the universe in order to have a gravitational field. If $$T_{\mu\nu} \,-\, \frac{1}{2} T g_{\mu\nu} \,\neq\, 0$$, then $$R_{\mu\nu} \,\neq\, 0$$. Consequently you can easily have a nontrivial gravitational field even with a constant (gradient-less) stress-energy density. Furthermore, if the stress-energy tensor has nonzero trace then the Ricci scalar is also necessarily nonzero regardless of the gradient of the stress-energy density.
This all follows from a simple and braindead inspection of the EFE, so if you say that it is a necessary condition for the gradient of the stress-energy density to be nonzero in order to have a gravitational field, you are explicitly contradicting the EFE.