How to find potential difference at A
Another approach is to take each one of the branches with batteries and substitute them with their Norton equivalent, i.e. a \$5\Omega\$ resistor in parallel with a \$ \dfrac{2V}{5\Omega}=0.4A\$ current source.







simulate this circuit &ndash; Schematic created using CircuitLab

Then you get three paralleled \$5\Omega\$ resistors driven by two paralleled \$0.4A\$ current sources. This means a total \$0.8A\$ flowing into an equivalent \$\dfrac{5\Omega}{3}\$ resistor. 







simulate this circuit

By Ohms'law you get:

\$ V_{AB} = \dfrac{5\Omega}{3} \times 0.8A = \dfrac{4}{3}V \approx 1.33V \$

EDIT

(Since it seems you have not grasped the concept of circuit ground).

A voltage is an electrical quantity always measured between two points in a circuit. Since it is cumbersome to draw schematics full of arrows (or +- symbols) specifying where are all the voltages you need, it is customary to take a common reference point (node, to be precise) for all voltages: that point (that node) is called circuit ground. Then you specify only node voltages, i.e. voltages between a node and the ground node. That is, for each node k, \$V_k = V_{kG} \$ where G indicates ground. This is a definition!

So, by marking a node (B in your case) with the ground symbol you are saying that any other node voltage is measured/specified with respect to that node (B). This allows specifying node voltages with only one subscript. Moreover this doesn't lead to a loss of generality, because it can be easily shown that a voltage \$V_{XY}\$ between any two points X and Y in the circuit can be written as a difference between two node voltages, i.e. \$V_{XY} = V_X - V_Y\$ (if either X or Y is ground, the corresponding node voltage is of course 0).