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.
I'll try and briefly run through some points (without mathematical detail) to see if this clears up any of your questions
• You seem to have misunderstood the 'blunder' part: the 'blunder' wasn't removing the cosmological constant, but adding it to his equations (in an ad-hoc way, at the time) in the first place.
• Today, dark energy isn't 'different' from the cosmological constant - the CC is just a possible (and the simplest) way to describe dark energy. (It's also the one that works best, despite the surrounding theoretical issues.)
• Einstein removed the CC because it was no longer needed for a static universe (and the other associated problems regarding stability, which you quoted).
Just to be clear, neither a cosmological constant nor any type of dark energy is needed for an expanding universe, but is needed for accelerated expansion. The universe was already expanding from the big bang. A universe where the CC is zero still expands.
Best Answer
As far as originally adding the term,
When Hubble published his paper in 1929 showing the universe to be expanding, Einstein reassessed:
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.