It is indeed possible to convert the system to an equivalent first order system, but let's first directly solve the given problem. The system's step response satisfies
$$y''(t)-y(t)=\delta(t)-u(t)\tag{1}$$
This implies that we can compute the step response of the given system by considering another system
$$y''(t)-y(t)=x(t)\tag{2}$$
and compute its impulse response ($x(t)=\delta(t)$) and its step response ($x(t)=u(t)$), and obtain the step response $(1)$ as the difference of the impulse response and the step response of the new system $(2)$.
For the step response we need to solve
$$y''(t)-y(t)=u(t)\tag{3}$$
Assuming the system is initially at rest, we get initial conditions $y(0)=y'(0)=0$. The homogeneous solution of $(3)$ is $y_h(t)=Ae^{t}+Be^{-t}$, $t>0$, whereas a particular solution is $y_p(t)=-1$, $t>0$. The complete solution is then
$$y(t)=y_h(t)+y_p(t)=Ae^{t}+Be^{-t}-1,\quad t>0\tag{4}$$
The constants $A$ and $B$ are determined by the initial conditions:
$$\left.\begin{align}y(0)&=A+B-1=0\\y'(0)&=A-B=0\end{align}\right\}\Rightarrow A=B=\frac12\tag{5}$$
Consequently, the step response of system $(2)$ is
$$y(t)=\frac12\left(e^t+e^{-t}\right)-1,\quad t>0\tag{6}$$
Its impulse response satisfies
$$y''(t)-y(t)=\delta(t)\tag{7}$$
and it can be obtained by differentiating the step response $(6)$:
$$y(t)=\frac12\left(e^t-e^{-t}\right),\quad t>0\tag{8}$$
Finally, according to $(1)$ and $(2)$, the step response of the original system is the difference between the impulse response and the step response given in $(8)$ and $(6)$:
$$y(t)=\frac12\left(e^t-e^{-t}\right)-\frac12\left(e^t+e^{-t}\right)+1=1-e^{-t},\quad t>0\tag{9}$$
Another approach would be to see that the original system can also be represented by a first-order system. Note that
$$y''(t)-y(t)=(y(t)+y'(t))'-(y(t)+y'(t))\tag{10}$$
So if $y''(t)-y(t)=x'(t)-x(t)$, it follows from $(10)$ that
$$y(t)+y'(t)=x(t)\tag{11}$$
With $x(t)=u(t)$ you obtain from $(11)$ the same step response as in $(9)$:
$$\begin{align}y_h(t)&=Ce^{-t}, &t>0\\
y_p(t)&=1, &t>0
\end{align}\\y(t)=y_h(t)+y_p(t)=Ce^{-t}+1,\quad t>0$$
From $y(0)=0$ we get $C=-1$, and the step response becomes
$$y(t)=1-e^{-t},\quad t>0$$
just like in $(9)$.
As a final check it is useful to see what happens when you solve the problem using the Laplace transform. The original difference equation transforms to
$$Y(s)(s^2-1)=X(s)(s-1)$$
which gives for the transfer function
$$H(s)=\frac{s-1}{s^2-1}=\frac{1}{s+1}$$
This shows that you have a pole-zero cancellation, and the system is actually a first-order system, just as described by Eq. $(11)$. For $x(t)=u(t)$ you get $X(s)=1/s$, and, consequently, for the Laplace transform of the step response
$$Y(s)=\frac{1}{s}\frac{1}{s+1}=\frac{1}{s}-\frac{1}{s+1}$$
which corresponds to the time domain function
$$y(t)=u(t)-e^{-t}u(t)=(1-e^{-t})u(t)$$
which is the same as $(9)$.
