Do non-commuting Hamiltonians have non-commuting time evolution operator? 
*

*If there are two Hamiltonians $H$ and $K$, $[H,K]\neq0$, do their time evolution operator, $e^{-iHt}$ and $e^{-iKt}$ for same time $t$ also non-commutative, $[e^{-iHt},e^{-iKt}]\neq0$, for all $t>0$?


*If it is not true for all time $t$, can we show this non-commutativity valid for at least for some time $t$?
I think this is true, by taking an expansion on exponential, but not sure this is rigorous or not.
 A: The short answer is yes the unitary evolution will not commute if considered in full generality of two parameter dependence but not under certain special circumstances (and also single parameter dependence). In fact it doesn't have to be two Hamiltonian although we will discuss that case towards the end. If we start with two Hermitian operators $\mathcal{O}_1(x_1)$ and $\mathcal{O}_2(x_2)$ and evolve one using the other then in the Heisenberg picture we will have,
$$i\hbar\frac{d\mathcal{O}_2}{dx_1}=[\mathcal{O}_2,\mathcal{O}_1]\tag{1}$$
Setting $\hbar=1$, in the unitary representation we can write this as,
$$O_2(x_1,x_2)=e^{i\mathcal{O}_1x_1}\mathcal{O}_2(x_2)e^{-i\mathcal{O}_1x_1}\tag{2}$$
Think of this as the global version of the more local statement of the Heisenberg equation. One way to prove an equivalence between the two is to Taylor expand around the origin. I like to think of this as zooming in on some one parameter space of the operator. What this should tell you is that if one operator has dependence on the evolution parameter of another operator (that is to say this operator is not conserved with respect to the other operator), then any new operator constructed from the commutator of these two operators should also not commute when considered in full generality. As an example consider,
$$\begin{align}
[e^{-i\mathcal{O}_1x_1},e^{-i\mathcal{O}_2x_2}]&=e^{-i(\mathcal{O}_1x_1+\mathcal{O}_2x_2-\frac{i}{2}[\mathcal{O}_1,\mathcal{O}_2]x_1x_2+\cdots)}-e^{-i(\mathcal{O}_1x_1+\mathcal{O}_2x_2+\frac{i}{2}[\mathcal{O}_1,\mathcal{O}_2]x_1x_2+\cdots)}\\
&=e^{-i(\mathcal{O}_1x_1+\mathcal{O}_2x_2)}\left(e^{-\frac{1}{2}[\mathcal{O}_1,\mathcal{O}_2]x_1x_2+\cdots}-e^{\frac{1}{2}[\mathcal{O}_1,\mathcal{O}_2]x_1x_2+\cdots}\right)
\end{align}$$
Note that I have used the Baker-Campbell-Hausdorff formula. We can see that unless,
$$e^{-\frac{1}{2}[\mathcal{O}_1,\mathcal{O}_2]x_1x_2+\cdots}=e^{\frac{1}{2}[\mathcal{O}_1,\mathcal{O}_2]x_1x_2+\cdots}\tag{3}$$
this commutator won't be zero. If I naively ignore the higher order terms (which will also be given in terms of the operator commutator), then this vanishing operator commutator would trivially satisfy this relation. For the higher order terms one should note that the odd order terms will not contribute to the exponential as they are symmetric. This comes from the fact that given,
$$e^Xe^Y=e^Z\implies Z(Y,X)=-Z(-X,-Y)$$
So in an expansion in the odd orders, we should get $X\leftrightarrow Y$ symmetry, and therefore I can cancel them out in $(3)$. Again even order terms will be antisymmetric under $\mathcal{O}_1\leftrightarrow\mathcal{O}_2$ and so $(3)$ would hold for all order. Additionally given that higher order terms depend on the second order commutator showing $(3)$ holds for second order, should be sufficient. Note that in higher orders there is a good chance that the overall commutation vanishes so we don't have to worry about them all that much.
Thus with this we come to another way of satisfying $(3)$, that is by demanding a periodicity constraint. Notice that the commutator of Hermitian operators will be imaginary. So if we consider a compact parameter space, for example $x_1\in S^1$ it is possible to satisfy $(3)$ whenever,
$$i[\mathcal{O}_1,\mathcal{O}_2]x_1x_2=2\pi n,\,n\in\mathbb{Z}\tag{4}$$
Just for the sake for playing around lets further consider the commutator $[p_x,L_y]=ip_z$ where I've again set $\hbar=1$ and the angular coordinate is periodic. Further let us be in the momentum state space, in which case some,
$$\theta_y=-2\pi n(xp_z)^{-1}$$
should satisfy $(3)$. Note that the periodic constraint would hold whenever the left hand side of $(4)$ stays real. Granted this is not exactly the best example but it should convey the idea.
Okay now you can consider time as the same parameter for both the operators. It can be shown that this would mean a contradiction with your assumption 1. First lets consider from $(1)$ the modification to $(4)$,
$$\begin{align}\frac{d\mathcal{O}_2}{dx_1}&=\frac{2\pi n}{x_1x_2}\tag{5}\\ \frac{d\mathcal{O}_1}{dx_2}&=-\frac{2\pi n}{x_1x_2}\tag{6}\end{align}$$
Then taking $x_1=x_2=t$,
$$\begin{align}\mathcal{O}_2+\mathcal{O}_1=f(t)\end{align}$$
would also satisfy $(3)$. This is also a direct consequence of $(1)$. However when we calculate the commutator from this,
$$\begin{align}
[\mathcal{O}_1,\mathcal{O}_2]=[t,\mathcal{O}_2]\frac{\partial f}{\partial t}&=-[t,\mathcal{O}_1]\frac{\partial f}{\partial t}\\
\implies i\frac{\partial f}{\partial t} &=0
\end{align}$$
So the function has to be constant (and real). Which tells us that this is some sort of total constant energy and also that the commutator vanishes. This should make sense since if there is a single time, the two Hamiltonians can be added to construct one conserved Hamiltonian. If the two Hamiltonians are not commutating, then you should be considering two separate time variables and solving $(5)$ and $(6)$ (along with the reality condition) should give you the answer for commutating evolution operators.
Finally as a reference you can check Howard Georgi's Lie Algebra in Particle physics.
A: Here is a counterexample to the first part of OP's question: Imagine $K$ is diagonalizable with  eigenvalue spectrum $\subseteq \mathbb{Z}$ within the integers. Then $e^{i2\pi K}={\bf 1}$ is the identity operator, which commutes with everything. So $[e^{i2\pi H},e^{i2\pi K}]=0$ commute even if $[H,K]\neq 0$ do not commute.
A: I assume that the non-commuting operators  $H$ and $K$ are selfadjoint and bounded to avoid issues with domains.
Hypothesis 1 is untenable. In general we cannot have that
$$[e^{itH}, e^{itK}]\neq 0$$
for all values of $t\neq 0$, because it is easy to construct elementary counterexamples as in @Qmechanic's answer.
Hypothesis 2 is valid in a more refined form.
Indeed, the only fact one can conclude from the non commutativity of $H$ and $K$ is that there is a sequence ${\mathbb R}\ni t_n\to 0$, as $n\to +\infty$, such that
$$[e^{it_nH}, e^{it_nK}]\neq 0\:.$$
In fact, if the sequence above did not exist, we would have an interval $I$ including $t=0$ where the one-parameter groups above commute. Taking the first (strong) derivative in $t\in I$,
$$[H e^{itH}, e^{itK}]+[e^{itH}, Ke^{itK}]=0\:. $$
Taking another $t$ derivative at $t=0$,
$$2[H,K]=0$$
that contraddicts the hypothesis.
