Einstein initially added the Cosmological Constant because (if I get this right) it seemed to him that the universe should be static. I agree that back then this would have been an obvious assumption. I'm curious now, before Hubble, where there any opinions/debates about whether the universe would be expanding or contracting?
[Physics] Why didn’t Newton have a cosmological constant
cosmological-constantcosmologygeneral-relativity
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This is a rather subtle question, which confused even Newton. It is very tempting to think that an initially static Newtonian universe with perfectly uniform mass density will not collapse, because the gravitational force cancels everywhere by symmetry. This is wrong.
Here's an analogous question: suppose a function $f$ obeys $$f''(x) = 1$$ and we want to solve for $f(x)$. Since every point on the real line is the same as every other point, we might think that by symmetry, $$f(x) = \text{constant}.$$ But this is completely wrong, because the second derivative of a constant is zero. And stepping back, the whole question doesn't make any sense, because there isn't enough information. To solve a general differential equation, you need boundary conditions.
One possible boundary condition is that the solution looks approximately even at infinity. That's enough to specify the solution everywhere, as $$f(x) = \frac{x^2}{2} + \text{constant}.$$ But now the translational symmetry has been broken: not every point is equivalent anymore, because we have a minimum at $x = 0$. This is inevitable. You can't solve the differential equation without boundary conditions, and any choice of boundary conditions breaks the symmetry.
Similarly in Newton's infinite universe we have $$\nabla^2 \phi = \rho$$ where $\rho$ is the constant mass density and $\phi$ is the gravitational potential, corresponding to $f$ in the previous example. Just as in that example, we "obviously" have by symmetry $$\phi(x) = \text{constant}$$ which indicates that the force vanishes everywhere. But this is wrong. Without boundary conditions, the subsequent evolution is not defined; it is like asking to solve for $x$ given only that $x$ is even. With any set of boundary conditions, you will have a point towards which everything collapses. So the answer to your question is that both the Newtonian and relativistic universes immediately start to collapse; the symmetry argument does not work in either one, so there is nothing strange to explain.
The reason this point isn't mentioned in most courses is that we often assume the gravitational potential goes to zero at infinity (in Newtonian gravity) or that the metric is asymptotically flat (in relativity). But this boundary condition doesn't work when the mass distribution extends to infinity as well, which leads to the pitfall here. The same point can lead to surprises in electrostatics.
We reasoned above in terms of potentials. A slightly different, but physically equivalent way of coming to the same conclusion is to directly use fields, by integrating the gravitational field due to each mass. In this case, the problem is that the field at any point isn't well-defined because the integrals don't converge. The only way to ensure convergence is to introduce a "regulator", which makes distant masses contribute less by fiat. But any such regulator, by effectively replacing the infinite distribution with a finite one, introduces a center towards which everything collapses; just like the boundary conditions, any regulator breaks the symmetry. So, again, collapse immediately begins.
In the end, both the Newtonian and relativistic universes immediately begin to collapse, and in both cases this can be prevented by adding a cosmological constant. In the Newtonian case, this is simply the trivial statement that $\nabla^2 \phi = \rho - \Lambda$ has constant solutions for $\phi$ when $\rho = \Lambda$. However, in both cases the solution is unstable: collapse will begin upon the introduction of any perturbations.
As far as originally adding the term,
It was a prejudice of the time that the universe was constant and eternal, forever unchanging — at least on the largest levels. This led Einstein to add a term to his initial equations in 1917. While the original formulation of general relativity included only the attractive form of gravity, this new term, called the cosmological constant, was a repulsive term. The attractive and repulsive forms of gravity could be tuned to balance one another, resulting in a stationary and unchanging cosmos.
When Hubble published his paper in 1929 showing the universe to be expanding, Einstein reassessed:
With the realization that his earlier prejudice for an unchanging cosmos was wrong, Einstein removed the cosmological constant from his equations. He was reported by physicist George Gamow as having called it his "biggest blunder."
Excerpts taken from Don Lincoln's Op-Ed.
On your clarification, allowing expansion in the original field equations was not his intention. Einstein was chasing after general covariance, and his underlying assumption of a static universe was apart from him arriving at this invariance of physical law. It was only when Friedmann and Lemaître found expanding solutions to his field equations that he modified it,
$$R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=T_{\mu\nu}\hspace{3mm}\Rightarrow\hspace{3mm}R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}+\Lambda g_{\mu\nu}=T_{\mu\nu},$$
sticking in the cosmological constant in order to permit steady-state solutions (the only ones he thought to be physical). However, the solutions to this modification were unstable, and a handful of years later Hubble came along to shed some light on the matter.
Best Answer
It's a very good question but the answer is that Newton's Universe actually doesn't have to expand or contract so no cosmological constant is needed. Well, it's a bit more subtle.
The right non-relativistic gravitational equation where one should add the vacuum energy density is the Poisson equation $$ \nabla^2 \phi_g = 4\pi G \rho + \Lambda_{{\rm Newton}}. $$ I added a Newtonian cosmological constant term. For a Newtonian cosmology with a uniform mass distribution at the cosmological scale (e.g. above hundreds of megaparsecs), you actually have to add this term (with a negative sign), to neutralize the mass density at the very long distance scale. If you omit this term, the $\phi_g$ potential has to have a Laplacian with a constant term, so $\phi_g$ itself will have to contain something like $\vec x^2$ which will inevitably be minimized at some point of the Universe - $\vec x =0$ in my conventions.
Amusingly enough, one may describe Newton's gravitational forces without any $\phi_g$, by manifestly summing the forces from other point masses in the Universe. It's an equivalent approach to calculate the acceleration but it allows us to avoid the problem with the preferred point in the Universe. I may just claim that the forces acting on the Earth that are caused by very distant objects cancel. This is equivalent to saying that the Earth is the $\vec x = 0$ point - and one may say the same thing for any other object in the Universe (a method to regulate the infrared divergences from the forces caused by very distant objects).
Needless to say, the assumption that we choose a "uniform cutoff" around every probe in the Universe is totally equivalent to adding the neutralizing Newtonian cosmological constant above. Also, you won't be able to invent any non-equivalent yet consistent Newtonian cosmologies with a uniform Universe but a nonzero cosmological constant. That's really because the Newtonian spacetime is flat and the cosmological constant is the curvature of the empty spacetime - which vanishes in Newton's theory by definition.