Of course it's possible. Rodrigues solved the problem in 1840.
In your somewhat idiosyncratic notation, $E_1=L_x$ and $E_2=L_y$ physicists learn in Sophomore year. I will, predictably, choose easy angles $\theta= \pi/2$, instead of your freaky $\theta=1$.
You then trivially have a roll and a pitch,
$$
\exp \left ({\pi\over 2} E_1 \right ) = \begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & 0
\end{bmatrix} , \qquad \exp \left({\pi\over 2} E_2 \right)
= \begin{bmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
-1 & 0 & 0
\end{bmatrix} ~~~\leadsto \\
\exp \left({\pi\over 2} E_1 \right) \exp \left({\pi\over 2} E_2 \right) =
\begin{bmatrix}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{bmatrix} \\ = \exp\left ( {2\pi\over 3} (E_1+E_2+E_3)/\sqrt 3\right ),
$$
Where we have used this formulation of the Rodrigues formula, for $\theta=2\pi /3$,
$$M\equiv E_1+E_2+E_3\equiv \sqrt{3} K=\begin{bmatrix}
0 & -1 & 1 \\
1 & 0 & -1 \\
-1 & 1 & 0
\end{bmatrix} , \leadsto\\
\exp (\theta K)= I+\sin\theta ~K+ (1-\cos\theta) K^2 , ~~~\leadsto \\
\exp \left ( {2\pi\over 3} M/\sqrt{3}\right ) =I +{1\over 2} M +{1+1/2\over 3} M^2.
$$
For general angle formulas, you should use the first link, the Rodrigues-Gibbs formula. (After you've appreciated its music, the Original Rodrigues paper is also instructive.)
We have the following first terms:
\begin{align}
Z(X,Y)& =\log(\exp X\exp Y) \\
&{}= X + Y + \frac{1}{2}[X,Y] +
\frac{1}{12}\left ([X,[X,Y]] +[Y,[Y,X]]\right ) \\
&{}\quad
- \frac {1}{24}[Y,[X,[X,Y]]] \\
&{}\quad
- \frac{1}{720}\left([Y,[Y,[Y,[Y,X]]]] + [X,[X,[X,[X,Y]]]] \right)
\\
&{}\quad +\frac{1}{360}\left([X,[Y,[Y,[Y,X]]]] + [Y,[X,[X,[X,Y]]]]\right)\\
&{}\quad
+ \frac{1}{120}\left([Y,[X,[Y,[X,Y]]]] + [X,[Y,[X,[Y,X]]]]\right)\\
&{}\quad
+ \frac{1}{240}\left([X,[Y,[X,[Y,[X, Y]]]]] \right)\\
&{}\quad
+ \frac{1}{720}\left([X,[Y,[X,[X,[X, Y]]]]] - [X,[X,[Y,[Y,[X, Y]]]]] \right)\\
&{}\quad
+ \frac{1}{1440}\left([X,[Y,[Y,[Y,[X, Y]]]]] - [X,[X,[Y,[X,[X, Y]]]]] \right) \\ &{}\quad + \cdots
\end{align}
Even if $[X,[X,Y]] = [Y,[X,Y]]$, the higher commutators do not vanish completely in general (find an example!). I don't see an easy finite sum.
Best Answer
Ok, so $exp(x)exp(y) = exp(x+y+ \frac{1}{2}[x,y] +\cdots )$ and $exp(y)exp(x) = exp(y+x+ \frac{1}{2}[y,x] +\cdots )$. Therefore, using $[y,x]=-[x,y]$, $$ [exp(x),exp(y)] = exp(x+y+ \frac{1}{2}[x,y] +\cdots ) - exp(x+y- \frac{1}{2}[x,y] +\cdots )$$ then, $$ [exp(x),exp(y)] = [x,y] +\cdots. $$ The answer would seem to fall out of the BCH relation for higher orders if you want higher orders.
An outline of how to continue: use BCH to expand as: (the right equality is meant to indicate $A$ is the terms in the argument of the exponential) $$ exp(x)exp(y) = exp(x+y+\frac{1}{2}[x,y]+\frac{1}{12}[x,[x,y]]-\frac{1}{12}[y,[y,x]] + \cdots ) =exp(A)$$ Likewise $$ exp(y)exp(x) = exp(x+y+\frac{1}{2}[y,x]+\frac{1}{12}[y,[y,x]]-\frac{1}{12}[x,[x,y]] + \cdots ) $$ Note, (the right equality is meant to indicate $B$ is the terms in the argument of the exponential) $$ exp(y)exp(x) = exp(x+y-\frac{1}{2}[x,y]-\frac{1}{12}[x,[x,y]] + \frac{1}{12}[y,[y,x]]\cdots ) = exp(B) $$ We would like to simplify the quantity $exp(A)-exp(B)$ order by order in $x,y$. Recall $exp(A) = 1+A+\frac{1}{2}A^2 + \cdots$ and $exp(B) = 1+B+\frac{1}{2}B^2 + \cdots$. Clearly the constant and first order terms in $x,y$ cancel. I think at second order everything cancels except $[x,y]$. At third order, I'm not sure without a lot more calculation, however, I hope the path is partly clear now, granted it is tedious. There is probably a better way. Perhaps a program of successive differentiation and evaluation.