What actually is the differential gain of an operational amplifier and why does its value change when we consider the common-mode gain?
$$\bbox[4px,border:1px solid red]{V_u=(1+\frac{R_2}{R_1})(v_+-v_-)}$$
The above assumption where you start from is actually wrong. 

\$(v_+-v_-)\$ is infinitesimaly small or zero ideally. 

The correct relation is:
$$V_u = a_d.v_d + a_c.v_c$$
$$ie.,V_u = a_d.(v_+-v_-) + a_c.\frac{(v_+ +v_-)}{2}$$
$$\bbox[8px,border:1px solid black]{V_u = G_{OL}.(v_+-v_-)} $$
Where \$a_d\$ or \$G_{OL}\$ is the open-loop differential gain or simply open-loop gain, and it is very big or infinite for ideal op amp. And open-loop common-mode gain \$a_c = 0\$.

Coming to closed-loop gains in your circuit:

$$V_u=(\frac{R_1+R_2}{R_1})V_1-(\frac{R_2}{R_1})V_2 \tag1$$

The circuit has a negative feedback and you can calculate closed-loop differential and common-mode gains using the relations:
$$V_1 = V_c+\frac{V_d}{2}$$
$$V_2 = V_c - \frac{V_d}{2}$$
where \$V_d\$ and \$V_c\$ are differential and common-mode components of \$V_1\$ and\$V_2\$:

$$V_d =V_1-V_2 $$
$$V_c = (V_1+V_2)/2 $$ 

Equation (1) can be simplified as:
$$V_u=(\frac{R_1+R_2}{R_1}).(V_c+\frac{V_d}{2})-(\frac{R_2}{R_1}).(V_c - \frac{V_d}{2})$$
$$\implies V_u = (\frac{1}{2}+\frac{R_2}{R_1}).V_d+1.V_c \tag2$$
Compare (2) with:
$$V_u = A_d.V_d+A_c.V_c$$
$$\bbox[8px,border:1px solid black]{\therefore A_d = (\frac{1}{2}+\frac{R_2}{R_1}), A_c = 1 }$$
where \$A_d\$ and \$A_c\$ are closed-loop differential and common-mode gains respectively.