No, an LQR controller (or trivially saturated LQR controller) will not give the same control signal as an MPC controller. You can (and typically want to) tune he MPC controller though such that it coincides with the LQR feedback once the system enters the state where the LQR feedback is and remains unsaturated.
If LQR gave the same control as MPC we would never use MPC, as MPC is several orders of magnitudes more computationally expensive.
You are dealing with multiple sources of errors. Namely, the ODE solver ode45 does not give an exact solution for $P(t)$ but a numerical approximation for it and the method you use to simulate the state is forward Euler, which is only a first order method. Since your system is unstable any perturbation might get amplified a lot. Also keep in mind the cost function that you are trying to minimize, because it might not be cost effective for finite horizon LQR to drive the state close to zero.
Fortunately one can improve the accuracy of many of the numerical results. Namely, there is an analytical solution for $P(t)$, and $x(t)$ can also be solved with ode45. The analytical solution for $P(t)$ can be found by using $P(t) = \bar{P} + V^{-1}(t)$, with $\bar{P}$ the stationary solution of $P(t)$ so by solving the related CARE. The dynamics of $V(t)$ can shown to be
$$
\dot{V} = V\,\mathcal{A}^\top + \mathcal{A}\,V - B\,R^{-1} B^\top,
$$
with $\mathcal{A} = A - B\,R^{-1} B^\top \bar{P}$. By using vectorization and the Kronecker product that dynamics can also be written as
$$
\dot{z} = M\,z,
$$
with $M = I \otimes \mathcal{A} + \mathcal{A} \otimes I$ and $z(t) = \text{vec}(V(t)) - M^{-1}\,\text{vec}(B\,R^{-1} B^\top)$. The solution for $z$ is given by
$$
z(t) = e^{M\,(t - \tau)}\,z(\tau).
$$
So $P(t)$, given $P(T)$, can be obtained using
\begin{align}
V(T) &= \left(P(T) - \bar{P}\right)^{-1}, \\
z(T) &= \text{vec}(V(T)) - M^{-1}\,\text{vec}(B\,R^{-1} B^\top), \\
V(t) &= \text{vec}^{-1}\!\left(e^{M\,(t - T)}\,z(T) + M^{-1}\,\text{vec}(B\,R^{-1} B^\top)\right), \\
P(t) &= \bar{P} + V^{-1}(t).
\end{align}
I also did some simulations using your system/script while changing only your numerical solution for $P(t)$ with the analytical solution, the integration method for the dynamics of the state from forward Euler to ode45 or both. From those results it does seem that using the analytical solution $P(t)$ seemed to have to biggest impact.
Best Answer
Depending on the prediction horizon and model dimensions MPC will be more or much more computationally expensive compared to LQR. This may lead to added delay or require more expensive hardware. Namely (infinite horizon) LQR just needs to compute a gain matrix beforehand and after that the computation cost online is fairly small.
It can also be noted that LQR can be used as a roll-out policy for MPC and there also exists finite horizon LQR, which has a time varying gain matrix (which can still be calculated in advance).