For a hydrogen-like ion, the energy levels can be given by the formula:

$$E_n = -\frac{Z^2}{n^2} E_0$$

where $E_n$ is the energy of the nth level, $Z$ is the atomic number (for Li$^{2+}$, $Z=3$), $n$ is the principal quantum number, and $E_0$ is the ground state energy of the hydrogen atom ($-13.6 \, eV$).

Given that state $S_2$ has energy equal to the ground state energy of the hydrogen atom and one radial node, we identify that $S_2$ corresponds to $n=2$ for a hydrogen atom. This is because for hydrogen-like species, the number of radial nodes is given by $n-1$, where $n$ is the principal quantum number. The ground state ($n=1$) has 0 nodes, the first excited state ($n=2$) has 1 radial node, etc.

Since the energy of $S_2$ is equal to the ground state energy of the hydrogen atom, we can directly compare the energies. For hydrogen ($Z=1$), the ground state energy ($n=1$) is:

$$E = -E_0$$

For the Li$^{2+}$ ion in state $S_2$ (which we established is equivalent to $n=2$ in terms of energy for hydrogen), we use the formula $E_n = -\frac{Z^2}{n^2} E_0$. Since $Z=3$ for Li$^{2+}$, and given that $S_2$ has the energy equivalent to the ground state of hydrogen ($E_0$), we solve for the energy ratio rather than the specific energy of $S_2$.

We then look at $S_1$, which we know must be the ground state for Li$^{2+}$ since it is the state before $S_2$ and has one radial node (indicating $n=2$ for $S_1$).

Thus, for $S_1$, which actually corresponds to $n=2$ for Li$^{2+}$, the energy in units of the hydrogen atom ground state energy is:

$$E_{S_1} = -\frac{Z^2}{n^2} E_0 = -\frac{3^2}{2^2}E_0 = -\frac{9}{4}E_0$$

Now, to express this in units of the hydrogen atom ground state energy ($-E_0$):

$$\frac{E_{S_1}}{E_0} = -\frac{9}{4} = -2.25$$

So, considering the provided options and the fact that energy levels are usually considered in positive values when comparing magnitudes, the correct answer is:

Option C: 2.25.