Special units in the $11$th cyclotomic field

Yes. Indeed $$ (1 + \zeta + \zeta^{10}) \, (\zeta + \zeta^4 + \zeta^7 + \zeta^{10}) = 1 $$ with $\sum_i a_i = 3$ and $\sum_i b_i = 4$; now change each $a_i$ to $a_i+3$ and each $b_i$ to $b_i+3$. (This is not the only solution: $(2+\zeta+\zeta^{-1})^{-1}$ also works with some room to spare.)

If I did this right there are a total of $1045 = 55 \cdot 19$ solutions, obtained from the following $19$ basic solutions by changing $a_i,b_i$ to $a_{ri+s\bmod 11}$ and $b_{ri+s \bmod 10}$ for all $s\bmod 11$ and $r=1,2,3,4,5$:

33433433433 | 33434343433
33434243433 | 33344344333
43424242434 | 34423532443
43234443234 | 34252525243
34251615243 | 33334543333
32344444323 | 42516161524
32434443423 | 52244344225
43244244234 | 54423132445
65421012456 | 42443334424
55331213355 | 44224542244
35234243253 | 35160706153
35160606153 | 33453135433
23253635232 | 35145154153
44214641244 | 12346564321
32463036423 | 33426162433
14642224641 | 45250505254
50274047205 | 34260706243
41615251614 | 15523532551
31608080613 | 23642324632

The solutions with $\sum_i a_i \zeta_i = 1+\zeta+\zeta^{10}$ and $(2+\zeta+\zeta^{10})^{-1}$ are in the first and fifth orbit respectively. Each orbit is of size $55$, not $11\cdot 10 = 110$, because every solution is fixed by some involution $(a_i,b_i) \leftrightarrow (s-a_i,b-a_i)$.

We seek units $\alpha := \sum_{i=0}^{10} a_i \zeta^i \in {\bf Z}[\zeta]$ and $\beta = \alpha^{-1}$ all of whose algebraic conjugates have absolute value at most $37$. (This condition is weaker than the required conditions that $a_i,b_i$ are nonnegative and sum to $36$ and $37$ respectively; we check this stricter condition at the end.) It is well known that every unit in a cyclotomic number field $F_N := {\bf Q}(e^{2\pi i/N})$ is a root of unity times a real unit; in our setting with $N=11$ this is equivalent to the observation that the $a_i$ and $b_i$ are symmetric under some reflection $i \leftrightarrow s-i$. I chose representatives that make the sequences $(a_i)$ and $(b_i)$ visibly symmetric.

Now let $v_j$ ($1\leq j\leq 5$) be the five embeddings into $\bf R$ of the real subfield $F_{11}^+ = {\bf Q}(\zeta+\zeta^{-1})$; and let $\lambda: (F_{11}^+)^* \to {\bf R}^5$ be the homomorphism $$ x \mapsto (\log|v_1(x)|, \log|v_2(x)|, \log|v_3(x)|, \log|v_4(x)|, \log|v_5(x)|). $$ The kernel of $\lambda$ is ${\pm 1}$, and the image of any element of norm $1$ is contained in the hyperplane $\{ (c_1,c_2,c_3,c_4,c_5) : \sum_{j=1}^5 c_j = 0 \}$. By the Dirichlet unit theorem, the group $U$ of units maps to a lattice in this hyperplane. For $F_{11}^+$ the units are "well-known"; we could also find generators for $U$, and thus for $\lambda(U)$, by consulting the LMFDB entry for $F_{11}^*$, or using the number-field routines of gp or similar packages.

We seek units satisfying the additional condition that $|c_j| \leq \log 37$ for each $j$. Thus they are lattice points in the sphere $$ \sum_{j=1}^5 c_j^2 \leq 5 (\log 37)^2 < 66. $$ I used the qfminim routine in gp to generate a list of all such lattice points; there are $5025$ nonzero $\pm$ pairs. For each one, I checked whether $\alpha = \sum_{i=0}^{10} a_i \zeta^i$ satisfies $A := \sum_i a_i \equiv 3 \bmod 11,$ and thus $B := \sum_{i=0}^{10} b_i \equiv 3^{-1} \equiv 4 \bmod 11.$ If so, then subtracting $\min_i a_i$ from each $a_i$ and $\min_i b_i$ from each $b_i$ makes the coefficients nonnegative. Then if $A \leq 36$ and $B \leq 37$, adding back $(36-A)/11$ to each $a_i$ and $(37-B)/11$ to each $b_i$ makes the sums exactly $36$ and $37$ respectively. Choosing one representative from each orbit yields the list of $19$ displayed at the start of this answer.