Is there a non-Hopfian lacunary hyperbolic group?

Since you say that you are also interested in other limits of hyperbolic groups, here is an example. It is not lacunary hyperbolic.

The first observation is that every f.g. group $N\rtimes\mathbf{Z}$ is a limit of groups $G_n$ that are HNN-extensions over f.g. subgroups of $N$. This is due to Bieri-Strebel (1978) and can be checked directly (it is used in the paper of Olshanskii-Osin-Sapir to provide an elementary amenable lacunary hyperbolic group). Now assume in addition that $N$ is locally finite. Then the $G_n$ are virtually free, hence are hyperbolic. This shows that any (locally finite)-by-cyclic f.g. group is a limit of virtually free (hence hyperbolic) groups.

Now here's a non-Hopfian example. Recall that Hall's group is the group of invertible $3\times 3$ triangular matrices with $a_{11}=a_{33}=1$. Consider Hall's group $H$ over the ring $A=\mathbf{F}_p[t^{\pm 1}]$, where $\mathbf{F}_p=\mathbf{Z}/p\mathbf{Z}$ and $p$ is a fixed prime.

Its center $Z(A)$ corresponds to matrices differing to the identity at the entry $a_{13}$ only. Set $B=\mathbf{F}_p[t]$ and $G=H/Z(B)$. Then it can be shown that $G$ is non-Hopfian. Indeed, the conjugation by the diagonal matrix $(t,1,1)$ restricts to an automorphism of $H$ mapping $Z(B)$ strictly into itself and thus induces a non-injective surjective endomorphism of $G$. Now $G$ is a limit of virtually free groups, by the previous argument.

I don't know how to adapt the construction to yield a lacunary hyperbolic (LH) group, but limits of hyperbolic groups are in general much more ubiquitous than LH groups, which demand refined constructions. As far as I understood, the various constructions of LH groups were performed in the literature are specially manufactured to yield LH groups, and I'm not aware of any group that was explicitly constructed independently, and was then shown to be a LH group.

A short answer is "why not?". A longer answer would be to look at the known examples of non-Hopfian groups and try to make them lacunary hyperbolic. A quite general construction can be found in our paper with Dani Wise (Sapir, Mark; Wise, Daniel T. Ascending HNN extensions of residually finite groups can be non-Hopfian and can have very few finite quotients. J. Pure Appl. Algebra 166 (2002), no. 1-2, 191–202.). See Lemma 3.1 there, in particular. It is quite possible that this construction or its slight modification can be lacunary hyperbolic.

Update 1.Another way to construct an example is the following. Start with the free group $F_2=\langle a,b\rangle$. Pick two words $U(a,b), V(a,b)$ satisfying small cancelation. That will be the non-injective surjective endomorphism. To make it non-injective, pick a word $W(x,y)$ and impose the relation $W(U, V)=1$. To make it surjective, pick two words $P(a,b), Q(a,b)$, and impose the relations $P(U,V)=a, Q(U,V)=b$. Now to make the map $a\to U, b\to V$ an endomorphism, for every relation $S(a,b)=1$ introduced already, we need to add the relation $S(U,V)=1$, then apply the same operation to the resulting presentation, etc. This defines an infinite presentation naturally subdivided into finite subsets. It remains to choose the words $U,V,W, P, Q$ so that each finite piece of the presentation defines a hyperbolic groups and the whole presentation is lacunary hyperbolic. Some kind of small cancelation theory may help here.

Update 2. Both constructions give limits of hyperbolic groups. To prove lacunar hyperbolicity one needs to estimate the growth of hyperbolicity constants $\delta$ vs the growth of the length of defining relations. The problem could be that the hyperbolicity constants of the intermediate groups grow too fast comparing to the lengths of relations. It needs to be checked in both cases. The lengths of relations grow exponentially fast but so do the hyperbolicity constants $\delta$. One needs to compare the bases of exponents. Fortunately, in the second construction, it seems to me that the base of exponent of the rels growth is approximately the maximal length of $U,V$. And the base of growth of $\delta$ is a constant that is independent of $U,V$ (say, $4$). But it needs to be checked.

Since Henry also discusses the property of being equationally Noetherian, I think the following observation is worth posting. And it is too long for a comment, so I post it as an answer.

The observation is that there exists a torsion free lacunary hyperbolic group that is not equationally Noetherian. In fact, such a group can be constructed directly. Alternatively, we can use the following theorem.

Theorem. Let $\mathcal H_n$ be the closure of the set of all $n$-generated torsion free non-cyclic hyperbolic groups in the space of marked group presentations. Then there exists a torsion free lacunary hyperbolic group $L$ such that the set of $n$-generator presentations of $L$ is dense in $\mathcal H_n$.

The proof uses the the same idea as in my paper mentioned by Henry. It is a bit too technical to be posted here (but it is just a page long and I did verify the details).

Now let us take the Ivanov-Storozhev non-hopfian group $G\in \mathcal H_2$. Since $G$ is not equationally Noetherian, there is a system of equations $S$ which is not equivalent to any finite subsystem on $G$. Let us assume that the coefficients in $S$ are written as words in generators, so we can consider the same system over $L$. Let $F$ be any finite subsystem of $S$.

Since $G$ is not equationally Noetherian, $S$ is not equivalent to $F$ over $G$. In particular, there is another finite subsystem $F_1$ of $S$ which contains $F$ and which is not equivalent to $F$ over $G$. Note that the property of a tuple of elements of $G$ to be a solution to a fixed finite system can be detected using some finite ball. Hence $F$ and $F_1$ are not equivalent over every group from some sufficiently small open neighborhood of $G$. In particular, $F$ is not equivalent to $F_1$ (and hence to $S$) over $L$. Thus we obtain

Corollary. The (torsion free lacunary hyperbolic) group $L$ is not equationally Noetherian.