Know a curve $γ(s)=(u(s), v(s))$ on a surface parametrized by arc length is a geodesic if and only if
$$\dfrac{d}{ds}(Eu'+Fv')=\dfrac{1}{2}(E_uu'^2+2F_uu'v'+G_uv'^2)$$
o
$$\dfrac{d}{ds}(Fu'+Gv')=\dfrac{1}{2}(E_vu'^2+2F_vu'v'+G_vv'^2).$$
Then, use the form
$$du^2+G(u,v)dv^2$$
If $K=0$ then $\sqrt{G}_{uu}=0$ so
$$G(u, v)=A(v)u+B(v).$$
But at $u=0$, $x_u$ and $x_v$ are unit so $B(v)=1$.
Also, the curve $u=0$ is a geodesic with $v$ arc length. So the geodesic equation
$$\dfrac{d}{ds}(Eu'+Fv')=\dfrac{1}{2}(E_uu'^2+2F_uu'v'+G_uv'^2)$$
gives $0=G_u(0, v)$ and this means in our case $A(v)=0$. The first fundamental
form is therefore
$$ds^2=du^2+dv^2$$
also is isometric to the plane.
It's pretty much a classical result that the catenoid and helicoid are isometric; if you look up a proof of that, you'll probably be able to discover the inconsistency in your reasoning.
One thought: there's no reason that the isometry has to take $x$ to $x$ and $y$ to $y$ in your chosen parameterizations, so your reasoning about the coefficients of dx and dy seems invalid. Although as I recall, in the standard catenoid-to-helicoid isometry, the radial coordinate on the helicoid ($x$) corresponds to the axial coordinate ($x$) on the helicoid, so maybe you're OK.
BTW, the function, for the catenoid, is $e^x + e^{-x}$; you may need some constants in there to make things work out.
BTW 2: there's an animation of the helicoid to catenoid isometry on the Wikipedia page for catenoids
More Detail
The problem is that an isometry is a map between surfaces that preserves fundamental forms. If you reparameterize the first surface, you have to reparameterize the second "in the same way" to still preserve the forms. But before you can do that, you need to find SOME isometry between SOME pair of parameterizations, and just picking two and hoping doesn't usually work. :(
Following the Wikipedia page, I find that two workable parameterizations are
\begin{align}
S_1(u, v) &= (\sinh v \,\sin u , \sinh v \,\cos u , u) \\
S_2(u, v) &= (\cosh v \,\cos u , \cosh v \,\sin u , v) \\
\end{align}
In your case, in the first parameterization, you have $x = \sinh v$, and in the second, $f(x) = \cosh v$. Since $\cosh^2 x = 1 + \sinh^2 x$, you have $f(x) = \sqrt{1 + x^2}$, just as you conjectured.
But the actual isometry from $S_2$ to $S_1$ is the map $(x, y) \mapsto (\sinh x, y)$.
So you need to compare the 1st fundamental form for $S_2$ at $(x, y)$ to the first fundamental form for $S_1$ at the point $(\sinh x, y)$. If you try that, things might work out.
Best Answer
Let $E$ be the plane defined by $z=0$. In your case you may consider $f\colon R\to R$ defined by $$f(x)=-\frac{1}{4}\left(\ln(-2x+\sqrt{1+4x^2})-2x\sqrt{1+4x^2}\right).$$ Verify that $f'(x)=\sqrt{1+4x^2}$. Define $\phi\colon S\to E$, $\phi(u,v,u^2):=\bigl(f(u),v,0\bigr)$. It's easy to verify that $\phi$ is an isometry; check the FFFs.
I've constructed $f$ in an obvious way.
Edit (much later): The construction of $f$ is obvious since the FFF of the given map is $$\begin{pmatrix}1+4u^2& 0\\ 0&1\end{pmatrix}.$$