Fake proof that $1$ is the solution of $x^2+x+1=0$

Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution?

Yes, exactly right. Substituting one equation in another has the potential to introduce more solutions, because you've thrown away information. Specifically, you threw away the original equations you were given. A solution to the original equation will be a solution to your new one, but the new one may have others.

For a silly example, consider $$ x + x = x + 4. $$ There is only one solution to the equation: $x = 4$. But suppose we substitute $x = 4$ on the left side for one of the $x$'s; then we get $x + 4 = x + 4$, or $4 = 4$, which has infinitely many solutions (every real number $x$).

In your example, you ended up with the equation $−x^2+\frac{1}{x}=0$, which is equivalent to the equation $-x^3 + 1 = 0$ (assuming $x \ne 0$), or $x^3 - 1 = 0$. This factors as $x^3 - 1 = (x - 1)(x^2 + x + 1)$ (as was pointed out in a comment), so you have introduced exactly 1 new solution.

Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution?

Yes. One important fact about mathematics is that when we manipulate an equation (or proposition, or ...), we can lose information along the way. E.g. from $$x^2=4$$ we can deduce $$0\cdot x^2=0\cdot 4,$$ and from that deduce $$0\cdot x^2=0.$$ $x=17$ is certainly a solution to this new equation, but not the original equation. Basically, we haven't done anything wrong - if $x^2=4$ then $0\cdot x^2=0$ - but we also haven't gotten an equivalent statement.

As a slightly less trivial example, from $$-y=4$$ we can deduce $$\vert -y\vert=\vert 4\vert,$$ or equivalently $$\vert y\vert=4;$$ $y=4$ satisfies this new equation, but not the original one. Again, we haven't made any mistakes, but we've lost information: the new equation $\vert y\vert=4$ tells us less than the old equation $-y=4$.

The equation you have ended up with is not the same as the original. In fact, multiplying each side by $-x$, you get:

$$ x^3 - 1 = 0 $$

And it is clear that $1$ is a solution of this equation. The way you ended up with this fact is that, as Lord Shark pointed out, we can factor $x^3 - 1 = (x-1)(x^2+x+1)$, and you essentially showed that:

$$ x^2 + x+1 = 0 \Rightarrow (x-1)(x^2+x+1)=0 $$

Which is pretty clear when you restate it like that.