Calculate gain of degenerated common-source stage using small-signal model
The current equation you show is the source current:

$$i_s=g_mv_1+\dfrac{v_{o}-v_s}{r_{ds}} $$

And \$i_s=\dfrac{v_s}{R_S}\$

So you have:

(1)
$$\dfrac{v_s}{R_s}=g_mv_1+\dfrac{v_{o}-v_s}{r_{ds}}$$ 

It's hard for me to follow what you did but you could use the equation that includes the current through \$R_d\$. That is,

(2)

$$ \dfrac{v_o}{R_d}+g_mv_1+\dfrac{v_o-v_s}{r_{ds}}=0$$

You could use either equation to start the process to find \$\dfrac{v_o}{v_i}\$

If you take equation 1, you can solve for \$v_o\$ to find:

$$ v_o=v_s\dfrac{r_{ds}}{R_s}+v_s-g_mv_1r_{ds}$$

As you noted, \$v_i=v_1+v_s\$, so the previous equation becomes (after some algebra):

(3)
$$v_o=v_s\bigg(\dfrac{r_{ds}+R_S+g_mr_{ds}R_s}{R_s}\bigg)-g_mr_{ds}v_i $$

Everything looks good except for the fact that I still have a \$v_s\$ term and need to get rid of it so that we can solve for \$\dfrac{v_o}{v_i}\$.

You can use the second equation (2), solve for \$v_s\$ and plug that in equation (3).

Solving for \$v_s\$ in equation (2) results in:

$$v_s=\dfrac{(R_d+r_{ds})v_o+g_mv_ir_{ds}R_d}{(g_mr_{ds}+1)R_d}$$

You still have to plug this value of \$v_s\$ in equation (3), but now you will have everything in terms of \$v_o\$ and \$v_i\$ so you should get the result you've been looking for.

This is not the only way to do this but it kind of follows the path you took.