Simple equation manipulation gives the wrong solution.
As I don't find the existing answers particularly illuminating, here's my own take on this.
There's no issue with your reasoning, except that you've proved something vacuous. To really drill this home, let's start by writing the argument in a simpler way:
Assume $x^2 + x + 1 = 0$.
It follows that $(x-1)(x^2 + x + 1) = 0$.
Hence $x^3 - 1 = 0$.
Thus $x^3 = 1$.
Therefore $x = 1$.
Now observe the following:
The above argument is directional. In particular, we've proved is that if our initial assumption of $x^2 + x + 1 = 0$ is true, then our final conclusion of $x = 1$ is true. We have not proved the converse.
The step from $x^3=1$ to $x=1$ forces us to work in the real line (notice that from $z^3 = 1$ we can't deduce $z = 1$, assuming that $z$ is a complex number).
Since we're working over the real line, the initial assumption of $x^2 + x + 1 = 0$ is false. To see this, compute a discriminant, or draw a graph.
By the principle of explosion, everything follows from a false assumption, including $x = 1$, $x=2$, and $x=-1/1893248129823489245894589$. So the above argument is logically correct, but tells us nothing about the relationship between the condition $x^2 + x + 1 = 0$ and the condition $x = 1$. There's no relationship between these equations that has somehow been established by our proof, despite that our proof is 100% correct.
I just want to emphasize that we haven't proved the back ward direction. That is, we haven't proved that if $x = 1$, then we can deduce that $x^2 + x + 1 = 0$. That's because of the step where we multiplied both sides by $(x-1)$. In general, it's a law of math that $x = y \rightarrow f(x) = f(y)$ for an arbitrary function $f$. So we can do the same thing to both sides of an equation, and the new equation will be a logical consequence of the old equation. In this case, the thing we're doing to both sides is multiplication by $(x-1)$. But there's no law of math that allows you to go backward, unless $f$ is an injective function. Exercise. Prove that there exists $y \in \mathbb{R}$ such that the function $y \mapsto (x-1)y$ fails to be injective.
There's also the question of what happens over the complex numbers. There's some suggestion here that we have to work over $\mathbb{C}$ to understand the false proof. I disagree with this. Because firstly, it's not a false proof, as long as you understand what's been proved. And secondly, the argument fares worse, not better, over the complex plane! As J. W. Tanner explains, it's not true that $x^3 = 1 \rightarrow x = 1$ over the complex plane. So we don't even get a chain of implications over $\mathbb{C}$, because the step from $x^3 = 1$ to $x = 1$ simply fails in that context. You might say hang on, we get some implications like this: $$\alpha \rightarrow \beta \leftarrow \gamma.$$ Where the motivation is that we're thinking of $\alpha$ as the statement $x^2 + x + 1 = 0$, and $\beta$ as the statement $x^3 = 1$, and $\gamma$ as the statement $x = 1$. So, maybe that tells us something about the relationship between $\alpha$ and $\gamma$? But in fact, logically speaking, having one implication go one way and the other go the other way tells us nothing about the relationship between $\alpha$ and $\gamma$.
The solutions to $x^2+x+1=0$ are solutions to $(x-1)(x^2+x+1)=x^3-1=0$.
There are $3$ solutions in $\mathbb C$ to $x^3-1=0$.
One is $x=1$, which does not satisfy $x^2+x+1=0$, and the other two are complex.
In any event, $x^3-1=0$ does not necessarily mean that $x=1$, contrary to what you wrote.