Why can one product in an equilibrium system not be produced without the other?
In the first experiment, the dissolved $\ce{CuSO4}$ reacted with $\ce{NH3(aq)}$ to form the soluble, deep blue/purple complex via equation 1, $\ce{[Cu(NH3)4]^{2+}}$, just as you said.  

1) $\ce{Cu^{2+} + 4NH3 &lt;--&gt; [Cu(NH3)4]^{2+}}$

Depending on the concentration of $\ce{NH3}$, you would also have precipitated some amount of insoluble $\ce{Cu(OH)2}$ from the hydroxide formed by the ammonia reacting with water as in equation 2:  

2) $\ce{NH3 + H2O &lt;--&gt; NH4+ + OH-}$  

The equilibrium of equation 2 lies far to the left, so when you add more ammonia, you aren't adding an equal amount of hydroxide.

The difference in the second experiment is that you explicitly added excess hydroxide.  In this case the high hydroxide concentration would have pushed the dominant product to $\ce{Cu(OH)2}$ via equation 3.  

3) $\ce{Cu^{2+} + 2OH- &lt;--&gt; Cu(OH)2}$ (solid)

Regarding the statement "The goal of the experiment was to see if one of the products could be produced without the other":  

The products of both experiments were the same, just to different degrees; in other words, under different pH conditions, different products predominated.  In the first experiment, you had far more ammonia than hydroxide, so equation 1 dominated; lots of $\ce{[Cu(NH3)4]^{2+}}$ and a small amount of $\ce{Cu(OH)2}$.  In the second experiment, you added excess hydroxide so lots of $\ce{Cu(OH)2}$ was formed an very little $\ce{[Cu(NH3)4]^{2+}}$.  

So the upshot is that these reactions are in equilibrium, and because of equation 2 you always had both $\ce{NH3}$ and $\ce{OH-}$ in your reaction mixture so that both equations 1 and 3 always produced some small amount of their respective products, $\ce{[Cu(NH3)4]^{2+}}$ and $\ce{Cu(OH)2}$, respectively.