Why #"Li"_2^+# is more stable than #"Li"_2# ?

No... it's the other way around. #"Li"_2# is more stable than #"Li"_2^(+)#, because the bond is (hypothetically) stronger (probably gas-phase).

Here we consider the molecular orbital diagram (MO) of #"Li"_2#:

The bond order can be calculated in a simple manner. Just take electrons that are in each MO, and

• for each electron in a bonding MO, it adds #0.5# to the bond order, because more bonding character strengthens the bond...
• for each electron in an antibonding MO, it subtracts #0.5# from the bond order, because more antibonding character weakens the bond...

Hence, the bond order of #"Li"_2# is

#1/2 + 1/2 - 1/2 - 1/2 + 1/2 = 1#,

indicating a single sigma bond (because they are in the #sigma_(2s)# MO, a sigma orbital).

On the other hand, #"Li"_2^(+)# has one less electron, and it must be from the #sigma_(2s)#, so it loses #0.5# of a bond order. Thus,

#color(blue)("Bond order" ("Li"_2^+) = 0.5)#

And so, it has, hypothetically, half of a sigma bond. Clearly, half of a bond is less stable than one entire bond of the same type.