Why can you use the one-sided laplace transform to solve differential equation describing a causal LTI-system?
If it is not equal to 0 for t&lt;0 how can we know we can take the one-sided laplace transform of the LHS?

Unfortunately, the way the problem is framed it is very difficult to do in a rigorous manner.  You can, by remembering that $0^-$ is shorthand for $-\epsilon$ as $\epsilon \to 0$, and taking limits everywhere, and having expressions where you're looking at intervals of $t$ from $-\infty$ to $-2\epsilon$ and other maddening things.
The easy way to do it is to separate the problem: solve the problem for the $x(t)$ that's given.  Then augment the problem by finding an $x(t)$ that is zero everywhere except in $0^- &lt; t &lt; 0^+$ that will result in $y(0^+) = 2$ (note the $0^+$ instead of $0^-$) and $\dot y(0^+) = 1$.  This involves setting $x(t)$ to a linear combination of $\delta(t)$ and $\delta^2(t)$ (as if $\delta(t)$ weren't wacky enough, $\delta^2(t) = d/dt\ \delta(t)$).
Then add the two solutions together.  It may not be what your prof had in mind, but it takes the least amount of hand-waving, and if you really want rigor you can break $\epsilon$ and put it to work.
By doing the above trick of expressing $x(t)$ as a linear combination of the specified $x(t)$ plus whatever sum of the Dirac impulse and it's derivatives, you cast the problem into one where the system is linear -- you're just using physically impossible values for $x(t)$.