Well, the thing is, you did find $k$ -- it just didn't pop out named $k$, so to speak. Everything you know about $k$ will be just as true for $ij$ here, it just looks a little different (but only superficially).
From the presentation, you know that $ij$ is in the group being presented, because it is a group, and must be closed under the group operation.
As Matt Samuel said in the comments, presentations aren't always nice - and often they're pretty demanding, computationally. But, with sheer willpower, you can often wrestle such a presentation to learn more about your group. Here's the way I would approach it, but I make no claims of efficiency!
Disclaimer: Working with presentations is like an exercise in using algebraic identities dozens of times; I find such a thing impossibly hard to follow looking at someone else's work, but moderately fun to do once or twice a year. So, you'll probably want to follow along with a pencil and paper, if you actually want to digest any of this.
Since $i^4 = 1$, we must have that $\langle i \rangle = \{1, i, i^2, i^3\}$ is a cyclic subgroup. And, since $i^2 = j^2$, we can verify that $\langle j \rangle$ is another cyclic subgroup of size four. Except, in addition, we can clean up by using a few relations, to see that $\langle j \rangle = \{1, j, j^2, j^3\} = \{1, j, i^2, i^2j\}$.
So far, we've identified six unique elements: $1, i, j, i^2, i^3,$ and $i^2j$. But, we haven't learned much about $ij$, so we'll at least find its order now. Note that by the relation $j^{-1}ij = i^{-1}$, we know that $iji = j$ (right multiply by $i$, and left multiply by $j$). Now
$$(ij)^2 = \underbrace{iji}_j\cdot j = j^2 = i^2,$$
and by the same reasoning above, we know that $\langle ij \rangle = \{1, (ij), (ij)^2, (ij)^3\} = \{1, ij, i^2, i^3j\}$.
So now our list of elements contains
\begin{array}{cccc} 1 & i & i^2 & i^3 \\ j & ij & i^2j & i^3j \end{array} at least. Now, in order to show that we've found everything, we'll have to show that multiplication (on either side) by $i$ or $j$ gives us back something from this list. Of course, left multiplication by $i$, and right multiplication by $j$, will easily be seen to give us back something from the list.
In general, it would be nice to be able to put all of our products in the form $i^mj^n$, at which point we'll show that $m \in \{0, 1, 2, 3\}$ and $n \in \{0, 1\}$, and our list is indeed exhaustive. In order to arrive at this form, the commutator $[j, i] = j^{-1}i^{-1}ji$ will be extremely helpful, as $ij[j, i] = ji$. We'll see that $[j,i]$ will allow us to let the $i$'s and $j$'s switch places if we're willing to pay the commutator price. We'll see that here, $[j, i] = i^2 = j^2 = (ij)^2$. Try it yourself, what follows is one way.
Since $j^{-1}i^{-1} = (ij)^{-1} = i^3j$, we have
$$j^{-1}i^{-1}ji = i^3jji = i^3j^2i = i^3i^2i = i^2.$$
In practice, this is quite useful. Let's pick $i^2j$ from our list above, and right multiply by $i$, attempting to simplify $i^2ji$ to give it the form $i^mj^n$. Since we know that $ji = ij[j,i] = iji^2 = ijj^2 = ij^3$, we have
$$i^2ji = i^2(ij^3) = i^3j^3 = i^{-1}j^2j = i^{-1}i^2j = ij,$$ something from our list.
I'll spare you the rest (verifying the list is truly exhaustive), because it's really more enlightening just to dig in and see what you can figure out. But hopefully it will give you more of an idea how such a thing can work, in practice. You could take it a step further and construct the multiplication table, or verify all the things you know about $Q_8$, using the element list here.
For some groups, presentations are really nice; you can basically collapse the multiplication table of the dihedral groups into a $2 \times 2$ table (reflections and rotations), using presentations.
Best Answer
Unfortunately, it is not quite enough.
What you have done so far is to construct a map $η\colon \{-1, i, j, k\} → \operatorname{SL}_2 \mathbb F_3$ such that $$η(i)^2 = η(j)^2 = η(k)^2 = η(i)η(j)η(k) = η(-1).$$ This is exactly how you start. By the universal property of group presentations there is now exactly one group morphism $$Q_8 = ⟨-1, i, j, k;~i^2 = j^2 = k^2 = ijk = -1⟩ → \operatorname{SL}_2 \mathbb F_3,$$ extending $η$. So far, so good. Now you need to know that this extension is an isomorphism.
However, this would also be possible if the target was the trivial group, not $\operatorname{SL}_2 \mathbb F_3$. So why should your particular map be an isomorphism onto its image?
To show that this extension is an isomorphism onto its image, you would need to still show that it is injective.
To show it’s injective, you can either try to argue that its kernel is trivial or to argue that $Q_8$ indeed only has at most eight elements, as suggested by Derek Holt. In both cases you need to examine the elements of $Q_8$ as strings in the letters $-1, i, j, k$ and use their relations.