## Return to Answer

2 added emphasis

I believe that the answer is NO. If you look at

Gutiérrez, Mauricio A.; Ratcliffe, John G. On the second homotopy group. Quart. J. Math. Oxford Ser. (2) 32 (1981), no. 125, 45–55.

Corollary 3 states that a "reduced 2-complex $K(X; R)$ is aspherical if and only if each element of $R$ is independent and not a proper power."

Now, "reduced" means that there is (a) only one 0-cell (true in your case), and the one cells represent distinct nontrivial elements of $\pi_1(K^1),$ where $K^1$ is the one-skeleton. Again seems to be true under your assumptions. $R$ are the relations (given by attaching maps of the 2-cells, I imagine), "independent" is too complicated to explain here (look at the paper), but in any case, the "not a proper power" condition is easy to violate.

EDIT Actually, independent is not too hard to explain. The definition is: a relator $r$ is independent if, setting $M$ to be the normal closure of $r,$ and $N$ the normal closure of $R - r,$ $M \cap N = [ M, N].$

As @Benjamin points out, above I am answering the complementary question, so to get the example that the OP wants, we need three independent elements in the free group on two generators which are not proper powers.

1

I believe that the answer is NO. If you look at

Gutiérrez, Mauricio A.; Ratcliffe, John G. On the second homotopy group. Quart. J. Math. Oxford Ser. (2) 32 (1981), no. 125, 45–55.

Corollary 3 states that a "reduced 2-complex $K(X; R)$ is aspherical if and only if each element of $R$ is independent and not a proper power."

Now, "reduced" means that there is (a) only one 0-cell (true in your case), and the one cells represent distinct nontrivial elements of $\pi_1(K^1),$ where $K^1$ is the one-skeleton. Again seems to be true under your assumptions. $R$ are the relations (given by attaching maps of the 2-cells, I imagine), "independent" is too complicated to explain here (look at the paper), but in any case, the "not a proper power" condition is easy to violate.
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