We proved that for any quadratic matrix $A \in \mathbb{R}^{n\times n}$, the derivative of $x(t) = e^{tA}$ is $x'(t) = Ae^{tA}$.
Suppose we have two matrices $A,B$ such that $e^{tA} = e^{tB}$ for all $t \in \mathbb{R}$. Then the derivatives of both functions must also be equal, i.e. $Ae^{tA} = Be^{tB} = Be^{tA}$. We then find $Ae^{tA}e^{-tA} = Be^{tA}e^{-tA}$ and since $e^{tA}e^{-tA} = I_n$, we obtain $A = B$.
This ought to show the injectivity of the matrix exponential, but I read that the matrix exponential is surjective (to the group of invertible matrices) but not injective. Pretty much all of the sites I visited involved things like Lie groups though and I'm still in my first year so I didn't understand much of what they were saying. Is the above proof correct or did I make any mistakes?
Summarizing the comments: your proof is fine, but it doesn't prove the injectivity of the matrix exponential. What it proves is that if $e^{tA} = e^{tB}$ for all $t$ then $A = B$, but for injectivity the hypothesis you want is $e^A = e^B$, and there's nothing to differentiate.