You're going to need to use Sylow's theorems.If $n_{11}$ is the number of Sylow $11$-subgroups of $G$ then $n_{11}\bigm|88$ and $n_{11}\equiv1\pmod{11}$ so $n_{11}=1$.Then $G$ has a normal Sylow $11$-subgroup $P_{11}$.Let $P_2$ be a Sylow $2$-subgroup of $G$.The recognition theorem for semidirect products shows that $G\cong P_{11}\rtimes_\varphi P_2$ for some group homomorphism $\varphi\colon P_2\to\text{Aut}(P_{11})$.
Recall that $\text{Aut}(P_{11})\cong(\mathbb{Z}/11\mathbb{Z})^\times\cong\mathbb{Z}/10\mathbb{Z}$ is cyclic of order $10$ and has a unique subgroup $K$ of order $2$ (i.e. the subgroup $\{\pm1\}\leq(\mathbb{Z}/11\mathbb{Z})^\times$).The image of $\varphi$ must lie in $K$.Let $N=\ker\varphi$.We know that $\varphi$ factors as the composition$$P_2\twoheadrightarrow P_2/N\hookrightarrow K\hookrightarrow\text{Aut}(P_{11}).$$Since $K$ has order $2$, there is only one choice for the injective homomorphism $P_2/N\hookrightarrow K$.Thus, the choice of $\varphi$ is determined by the choice of a normal subgroup $N$ of $P_2$ of index $1$ or $2$.It remains to go through the possibilities for $P_2$ and the possible normal subgroups of $P_2$ of index $1$ or $2$.
Two notes:
(1) Subgroups of index $1$ or $2$ are necessarily normal, so requiring normality is redundant.
(2) If $N$ and $N^\prime$ differ by an automorphism $\sigma$ of $P_2$ (meaning that $N^\prime=\sigma(N)$) then the resulting groups $G=P_{11}\rtimes_\varphi P_2$ and $G=P_{11}\rtimes_{\varphi^\prime}P_2$ will be isomorphic.Thus, we only need to look at one $N$ from each automorphism-class.
Case 1: $P=C_8$. There are two possibilities for $N$: $C_4$ and $C_8$.
Case 2: $P=C_2\times C_4$. There are three possibilities for $N$: $C_2\times C_2$, $C_4$, and $C_2\times C_4$.
Case 3: $P=C_2\times C_2\times C_2$. There are two possibilities for $N$: $C_2\times C_2$ and $C_2\times C_2\times C_2$.
Case 4: $P=D_4$. There are three possibilities for $N$: $C_2\times C_2$, $C_4$, and $D_4$.
Case 5: $P=Q_8$. There are two possibilities for $N$: $C_4$ and $Q_8$.
This gives $12$ groups.
There are a few points to make though:
(1) You need to check that I really did cover every automorphism-class of index $1$ or $2$ subgroups of $P_2$.This is a pain but it's just a matter of running through what the automorphism groups of each possible $P_2$ do to its subgroups of index $2$.
(2) You need to check that these $12$ groups are pairwise non-isomorphic.This isn't so bad. There is no overlap between the five cases because each case has a different Sylow $2$-subgroup.Also, there is no overlap between picking $N=P_2$ (which results in $G$ having a normal Sylow $2$-subgroup) and picking $N\lneq P_2$ (which results in $G$ not having a normal Sylow $2$-subgroup). Then there are just two pairs of groups which you have to check are not isomorphic (which can be done by counting elements of 2).
(3) We know many of these groups. In the cases where $N=P_2$, we obtain the direct products
$C_{11}\times C_8\cong C_{88}$.
$C_{11}\times C_2\times C_4\cong C_{44}\times C_2$.
$C_{11}\times C_2\times C_2\times C_2\cong C_{22}\times C_2\times C_2$.
$C_{11}\times D_4$.
$C_{11}\times Q_8$.
In the cases where $N\lneq P_2$, we can still identify most of the groups:
$C_{11}\rtimes(C_2\times C_4)$ with $N=C_4$ is isomorphic to $D_{11}\times C_4$.
$C_{11}\rtimes(C_2\times C_2\times C_2)$ with $N=C_2\times C_2$ is isomorphic to $D_{11}\times C_2\times C_2\cong D_{22}\times C_2$.
$C_{11}\rtimes D_4$ with $N=C_4$ is isomorphic to $D_{44}$.
$C_{11}\rtimes(C_2\times C_4)$ with $N=C_2\times C_2$ is isomorphic to $\text{Dic}_{11}\times C_2$.
$C_{22}\rtimes Q_8$ with $N=C_4$ is isomorphic to $\text{Dic}_{22}$.
$C_{11}\rtimes C_8$ is the only one that I can't name.