Why does cancelling change this equation?

When you cancel out $t$ on both sides you are assuming $t\neq0$.

To be rigorous you need to divide into two cases,

1. $t\neq0$, then you can cancel out $t$.
2. $t=0$, then you cannot cancel out $t$ as division by $0$ is not allowed.

I think the answer lies in the steps you ignored, that is, the "cancelling out" process:

"Cancelling out" with few extra intermediate steps usually ignored:

$$2t^2+t^3-t^4=0\implies t^2(2+t-t^2)=0$$

$$\implies t^2=0\text{ or }(2+t-t^2)=0$$.

You must guard against division by zero. When you "cancel out $t^2$", you are dividing both sides of your equation by $t^2$. Consequently, the correct form of inference is

$$\begin{align*} 2t^2+t^3 &= t^4 & & \\ 2 + t &= t^2 \qquad \text{ or } \qquad t^2 = 0. \end{align*}$$

The resulting left choice has the solution set $\{-1,2\}$ and the resulting right choice has the solution set $\{0,0\}$, which together are the solution set of the first equation.

Note that the same thing happens in reverse. Multiplying both sides of an equation by zero can result in craziness. We can agree that $1 \neq 2$, but this does not mean $0 \cdot 1 \neq 0 \cdot 2$ because both sides of this are zero. It can be harder to see when one is doing this when using a more complicated expression that is only sometimes zero. For instance,

$$\begin{align*} 2 + t &= t^2 \\ t^2(2 + t) &= t^4 \qquad \text{ and } \qquad t^2 \neq 0 \\ 2t^2+t^3 &= t^4 \qquad \text{ and } \qquad t^2 \neq 0 \end{align*}$$

The left choice has solution set $\{-1,0,0,2\}$ and the right choice has solution set $\{0,0\}$. The (set) difference of these is $\{-1,2\}$, the solution set of the first equation.

None of this is hard to see when multiplying or dividing by constants. Either the constant is nonzero and everything works or the constant is zero and everyone can see that something bogus is going on. However, when we're not just using constants, a little more care is needed.

Write your equation as $$t^4 -t^3 - 2t^2 = 0,$$ factoring out $t$ to get $$t^2(t^2 - t - 2) = 0.$$ Factoring once more, we have $$t^2(t- 2)(t+1) = 0$$ so the roots are $-1,0,0,2$.

If you leave out the $t^2$ factor you lose the $0,0$ roots.

No such thing as cancelling actually exists. What you do in those moments is to divide both sides of an equation by the same value. This means, cancelling like this:

$2t^2 + t^3 = t^4$ to $2t + t^2 = t^3$

Is actually a shortcut for:

$2t^2 + t^3 = t^4$ to $(2t^2 + t^3)/t = t^4/t$ to $2t + t^2 = t^3$

And you can only do that by assuming $t \neq 0$, so you discard that solution by doing such operation. Perhaps $0$ would still be a valid solution if the exponents are $> 2$ in each term (so $0$ would still be a valid root for them) but you should never assume that.