Have a look at Hartshorne's Geometry: Euclid and Beyond. He uses Hilbert's axioms for geometry and discusses (section 11) the following "circle-circle intersection axiom (E)":
Given two circles $\Gamma,\Delta$, if $\Delta$ contains at least one point inside $\Gamma$, and $\Delta$ contains at least one point outside $\Gamma$, then $\Gamma$ and $\Delta$ will meet.
Note that the hypothesis of this axiom is (at least arguably) a different way to say "unless they obviously don't" which is nevertheless based on betweenness and order rather than distance.
Hartshorne shows (section 12) that essentially all of Euclidean geometry can be developed in a plane satisfying the two-dimensional version of all of Hilbert's axioms, but with the completeness axiom replaced by the above axiom (E). This answers your second question: completeness is not needed, all you need is (E). Finally, in section 16 Hartshorne shows that your intuition about the algebraic version of (E) is correct: the cartesian plane over an ordered field satisfies (E) iff all positive numbers have square roots.
Here is an expanded version of my previous comments. There are a lot of things to check here and I haven't.
In my opinion the right axioms for Euclidean/Hyperbolic geometry are the Tarski axioms. Tarski works in a system where the domain is $\mathbb{R}^2$ and you have two relations, a ternary betweenness relation and an equidistance relation $E(x,y;x',y')$ which says that the line segments $\overline{xy}$ and $\overline{x',y'}$ are congruent. In this system all the axioms are phrased in terms of points, not in terms of both points and lines like in Euclid. Lines emerge as definable sets.
The Tarski axioms for Euclidean geometry are on Wikipedia, and you just change the parallel postulate to get axioms for Hyperbolic geometry. The usual proof of bi-interpretability goes by showing that both theories are bi-interpretable with the theory of $(\mathbb{R},+,×,<)$, i.e. the theory of real closed fields. I'm not sure where the hyperbolic case was written down, maybe by Szmielew. I don't know of a synthetic version of the proof.
But it should be possible to get a synthetic proof. There are well-known Euclidean models of the hyperbolic plane, like the Klein model, so it should be enough to make a Hyperbolic model of the Euclidean plane.
Edit: I think that my initial attempt at the Hyperbolic model of the Euclidean plane fails, because equidistance isn't definable. Following a suggestion of @ColinMcLarty, I will describe a different model that I think works. This model is from Greenberg's book "Euclidean and Non-euclidean geometries".
Let $\mathbb{H}$ be the Hyperbolic plane. In this model the points are just the points in $\mathbb{H}$ and the lines all all lines in $\mathbb{H}$ through the origin together with the curves in $\mathbb{H}$ that are equidistant from a line through the origin.
To get a model of Euclidean geometry we need a betweenness and equidistance relationship, and these relationships need to be definable in $\mathbb{H}$. The betweenness relationship is easy, you just need to observe that the lines form a uniformly definable family of sets. Equidistance is more complicated.
Greenberg describes a map $\rho : \mathbb{H} \to \mathbb{R}^2$ and says that the equidistance relation on $\mathbb{H}$ is the pull-back of the equidistance relation on $\mathbb{R}^2$ by $\rho$. We can put polar coordinates on $\mathbb{H}$ in the same way as on the Euclidean plane. We fix a ray $\ell$ through the origin and let the polar coordinates of $p \in \mathbb{H}$ be $(r,\theta)$ where $r$ is the Hyperbolic distance from the origin to $p$ and $\theta$ is the angle that $p$ makes with $\ell$. If $p \in \mathbb{H}$ is the point with polar coordinates $(r,\theta)$ then $\rho(p)$ is the point
$$
\rho(p) = (\sinh r \sin \theta, \sinh r \cos \theta) = \sinh r (\cos \theta, \sin \theta).
$$
So $\rho(p)$ is the point in $\mathbb{R}^2$ whose (euclidean) polar coordinates are $(\sinh r, \theta)$.
Now let's suppose that $\mathbb{H}$ is the Poincare disc model. I think it is enough to show that $\rho$ is semialgebraic, i.e. definable in $(\mathbb{R},+,\times,<)$. The Hyperbolic distance $r$ of $p$ from the origin is not a semialgebraic function of $p$, but it is the arcsinh of a semialgebraic functions, see Wikipedia. (On Wikipedia there is a factor of $2$ in from the arcsinh, but that is just a scaling factor, so we can drop it). I think it should be pretty clear that $\rho$ is semialgebraic.
Once we know that $\rho$ is semialgebraic we know that the equidistance relation on our Hyperbolic model of Euclidean geometry is semialgebraic. This should imply that it is definable in Hyperbolic geometry, but one would need extra arguments to see it directly.
Best Answer
Here is one way to construct an equilateral triangle on $p$ and $q$ using Tarski's axioms -- i.e. combining the existential axioms to get the vertex of such a triangle, omitting the proof by the other axioms that this works.
Let $vw\ge xy$ abbreviate $\exists z(Bxyz \wedge vw = xz)$.
By Lower Dimension, there are $a,b,c$ which are not collinear. Assume without loss of generality that $c$ is not on the line between $p$ and $q$, and that $cp \ge cq$.
By Continuity, choose $m$ with $Bpmq$ and $mp=mq$.
By Continuity, choose $d$ with $Bcdp$ and $dp=dq$.
By Segment Construction, choose $e$ with $Bmde$ and $de=pq$.
By Continuity, choose $t$ with $Bmte$ and $pt=pq$.
For all these uses of Continuity, the axiom actually requires a pair of inequalities rather than an equality. So for the last line, we can let $\phi(x) = Bmxe \wedge px \le pq$, and $\psi(y)=Bmye \wedge py \ge pq$, and use the continuity axiom in the form $$\forall x\, \forall y\,[(\phi(x) \wedge \psi(y)) \to Bmxy]\, \to\, \exists t\, \forall x\, \forall y\,[(\phi(x) \wedge \psi(y)) \to Bxty]$$ The other uses of the continuity axiom are similar.