For what $X \in \mathfrak{su}(N)$ do we have $e^{2\pi X}=1$? The Lie algebra $\mathfrak{su}(N)$ consists of skew-Hermitian $N\times N$ matrices. It has an $\mathbb R$-basis $\{X_{ij}, Y_{ij}\}_{1\leq i<j\leq N} \cup\{Z_i\}_{1\leq i \leq N-1}$, where $X_{ij}$ has $(i,j)$-entry 1, $(j,i)$-entry -1, other entries 0, and $Y_{ij}$ has $(i,j)$- and $(j,i)$-entry $i$, other entries 0. Finally, $Z_i$ has $(i,i)$-entry $i$ and $(N,N)$-entry $-i$.
For simplicity, denote the above generators as $\{A_i\}$.
One can directly compute
$$e^{2\pi X_{ij}} = e^{2\pi Y_{ij}}=e^{2\pi Z_i}=1.$$
For $N=2$, this can be understood using the "almost" isomorphism between $SU(2)$ and $SO(3)$, since $2\pi$-rotation takes back to the original position. (My convention of generators are the twice of the one conventionally used.)
Furthermore, at least for $N=2$, we also have $e^{2\pi \xi_jA_j}=1$ (Einstein summation applied) for all $\xi_j \in \mathbb R$ with $\sum \xi_j^2=1$. In terms of $SO(3)$, this can be understood as a $2\pi$-rotation with respect to a general axis.
Question: Is this result generalizes to arbitrary $N$? More precisely, do we have
$$e^{2\pi \xi_jA_j}=1$$
for all $\xi_j \in \mathbb R$ with
$$\sum \xi_j^2=1\;?$$
I ask this question since for general $N$ we don't have an "almost isomorphism" with the rotation group.
 A: I don't know how to answer the title question, but I can prove an answer of "no" to the question in the body of your post.  Namely, for each $N\geq 3$, there are basis elements $A_1$ and $A_2$ and $\xi_1,\xi_2$ with $\xi_1^2 + \xi_2^2 = 1$ for which $e^{2\pi(\xi_1 A_1 + \xi_2 A_2)}$ is not the identity.
Let $A_1 = Z_1$ and $A_2 = Z_2$.  Let $(\xi_1,\xi_2)$ be any pair whose squares sum to 1 and for which neither $\xi_i$ is an integer.  Said another way, pick any $(\xi_1, \xi_2)$ on the unit circle in $\mathbb{R}^2$, except for $(0,\pm 1)$ and $(\pm 1,0)$.
Noting that $Z_1$ and $Z_2$ commute, it follows that $e^{2\pi(\xi_1 Z_1 + \xi_2 Z_2)} = e^{2\pi \xi_1 Z_1}e^{2\pi \xi_2 Z_2}$.
Because both $Z_i$ are diagonal, the exponential is easy to compute: we get the diagonal matrix whose diagonal entries are the exponentials of those in $2\pi \xi_i Z_i$.
Thus, \begin{align*} e^{2\pi(\xi_1 Z_1 + \xi_2 Z_2)} &= \operatorname{diag} \operatorname{diag}( e^{2\pi \xi_1 i}, 1,..., e^{-2\pi \xi_1 i}) \operatorname{diag}(1, e^{2\pi xi_2 i},..., e^{-2\pi \xi_2 Z_2})\\ &= \operatorname{diag}(e^{2\pi \xi_1 i}, e^{2\pi \xi_2 i},..., e^{-2\pi i (\xi_1 + \xi_2)})\end{align*}
By hypothesis, $\xi_1$ is not an integer, so $e^{2\pi \xi_1 i}\neq 1$.
The same argument shows that any compact Lie group of rank $\geq 2$ has an answer of "No" for any choice of basis.  (On the other hand, as mentioned in your post, the answer is "yes" for compact rank 1 groups.  However, there are only three such connected examples:  $SU(2), SO(3)$ and $S^1$.)
