The gravitational force would decrease at the point $B$ and increase at the point $A$. Actually this happens at all distances, and it's what causes the tides. Moving the Moon in until it touches the Earth is just an extreme case.
As long as the gravitational fields are weak (in this case weak means a lot less than a black hole) you can simply add the gravitational forces from the Earth and the Moon.
At point $B$ the total force is given by:
$$ F_B = F_E - F_{MB} $$
because the two forces point in different directions and oppose each other. So $F < F_E$ and the gravitational field is less than the field of the Earth alone. At point $A$ the total force is given by:
$$ F_A = F_E + F_{MA} $$
because the two forces point in the same direction and reinforce each other. So $F > F_E$ and the gravitational field is greater than the field of the Earth alone.
We can put numbers on this. I'll calculate the acceleration, which is the force per unit mass. At Earth's surface this is 1g, i.e. $9.81$ m/s$^2$, so $F_E = 9.81$ m/s$^2$. For a mass $M$ at a distance $r$ the acceleration is simply:
$$ a = \frac{GM}{r^2} \tag{1} $$
To calculate $F_{MB}$ we have to set $M$ equal to the mass of the Moon ($7.35 \times 10^{22}$ kg) and $r$ equal to the radius of the Moon ($1.74 \times 10^6$ m), and using equation (1) the acceleration is:
$$ a \approx 1.62 \text{m/s}^2 $$
To calculate $F_{MA}$ we set $r$ to the radius of the Moon plus the diameter of the Earth ($1.28 \times 10^7$ m), and using equation (1) the acceleration is:
$$ a \approx 0.023 \text{m/s}^2 $$
Combining these results and converting the accelerations to $g$ we get:
Point B: 0.835g
Point A: 1.002g