Proton: 2 up, 1 down quark, Neutron: 2 down, 1 up, how can Neutron: proton + electron?

A neutron is not "a proton and an electron". A neutron is not composed of a proton and an electron inside of the neutron.

In quantum mechanics, particles can appear and disappear or change into other particles. With the neutron, one of the down quarks can decay change into an up quark by emitting a W boson, turning into a proton. The W boson quickly decays into an electron and an electron antineutrino. The new up quark didn't exist until the down quark turned into it. The W boson is what is called a virtual particle. It doesn't exist in the classical sense, it's just kind of there in the ambiguous region of spacetime where the decay occurs. The electron and antineutrino didn't exist until the decay.

Here is a Feynman diagram of the process, from here:

A neutron isn't a proton and an electron.

The reaction involved in beta decay is $$n \to p + e^- + \bar{\nu}_e$$ where $\bar{\nu}_e$ is an electron anti-neutrino.

But even that doesn't mean a neutron is a proton plus an electron plus a anti-neutrino. It means that a neutron's quantum numbers are the same as a state consisting of a proton an electron and an electron anti-neutrino with the proper angular momentum relationship. And that plus the fact that a neutron's mass energy exceeds that of the products means that the decay is both allowed and mandatory.

The other answers are right, but I'd like to offer the opposite statement, too: A neutron is a proton plus an electron.

One wonderful thing about nuclear and particle physics is that you can do "arithmetics" with particles. In some sense, you can say a proton and an electron add up to a neutron, $p + e^{-}= n$. This is what you have in beta decay (if you are missing something, bear with me). But you can also "do the same on both sides" like in math (aka equivalence transformations). Lets add a positron:

$$\begin{aligned} n &= p + e^{-} \qquad &\big|_{+e^{+}} \\ \Rightarrow \quad n + e^{+} &= p + e^{-} + e^{+} = p & \end{aligned}$$

Note that the positron and the electron cancel out. The resulting formula $p = n + e^{+}$ can also happen in nature, and is called beta-plus decay. You can do all kinds of transformations, for example subtracting particles (which is the same as adding antiparticles). You can also turn an electron into a muon by removing an electron neutrino and adding a muon neutrino. I like to think of it as removing the electron-ness, and adding muon-ness:

$$e - \nu_{e} + \nu_{\mu} = \mu$$ or more conventionally: $$e + \bar\nu_{e} + \nu_{\mu} = \mu$$

This calculus works on the scale of nuclei, nucleons, and even quarks. This is the Feynman diagram from Johnathan Gross' answer:

$$\begin{aligned} p &= u+u+d \\ &= u+d+(u+W^{-}) \\ &= u+d+u+(\bar\nu_e+e^-) \\ &= n + \bar\nu_e+e^- \end{aligned}$$

Now we see that the formula in the beginning is incomplete, we were missing the neutrino. The reason it seemed to work without is that we were only considering the electric charge, but the neutrino is electrically uncharged.

The reason these cute calculations work is essentially a property called crossing symmetry, and the fact that quantum numbers are preserved. I think of this like a beginners version of Feynman diagrams (and in fact I think this is usually taught first. I figured this out in school, in the context of nuclear decay, and this was a major "wow" moment that increased my interest in particle physics.)

Of course, there are some downsides to this simplistic view. The most important is that there is no consideration of masses, and mass defects. Only heavier particles can decay into lighter ones, plus energy. But apart from that, this "calculating with particles" can be very useful, for example if you forgot whether to put a neutrino or an anti-neutrino.