If $(-1) \cdot (-1) = +1$ shouldn't $(+1) \cdot (+1) = -1$? The common multiplication rules are 
$$(-1) \cdot (-1) = +1 \\ 
(+1) \cdot (+1) = +1$$
But these rules seem asymmetric. Because of these rules it is not possible e.g. to solve the equation
$$x^2 = -1 $$
without introducing complex numbers. There would be two "more symmetric" alternatives:
$$(-1) \cdot (-1) = -1 \\ 
(+1) \cdot (+1) = +1$$
or
$$(-1) \cdot (-1) = +1 \\ 
(+1) \cdot (+1) = -1$$
Both alternatives would allow to solve the equations
$$x^2 = +1 \\ x^2 = -1 $$
without the need to introduce complex numbers. However the two more symmetric alternatives are not used. What is the reason for that? Would math break down, if one would adopt one of these different conventions?

Math would certainly break down, here's why. Assume that every number is equal to itself, addition is associative, multiplication distributes over addition , $1$ and $-1$ are additive inverses, $1$ is the multiplicative identity, and $0$ is the additive identity. Then it follows that:
$$\begin{array}{lll}
\bigg(1\cdot 1 + 1\cdot (-1)\bigg) + (-1)\cdot(-1) &=& \bigg(1\cdot 1 + 1\cdot (-1)\bigg) + (-1)\cdot(-1)\\
\bigg(1\cdot 1 + 1\cdot (-1)\bigg) + (-1)\cdot(-1) &=& 1\cdot 1 + \bigg(1\cdot (-1) + (-1)\cdot(-1)\bigg)\\
1\bigg(1+(-1)\bigg) + (-1)\cdot(-1) &=& 1\cdot 1 + (-1)\bigg(1 + (-1)\bigg)\\
1\bigg(0\bigg) + (-1)\cdot(-1) &=& 1\cdot 1 + (-1)\bigg(0\bigg)\\
0 + (-1)\cdot(-1) &=& 1\cdot 1 + 0\\
(-1)\cdot(-1) &=& 1\cdot 1 \\
\end{array}$$
The only loose end not assumed above is  $(-1)\cdot 0 = 0$
$$(-1)\cdot 0 = (-1)(0+0)$$
$$(-1)\cdot 0 = (-1)\cdot 0+ (-1)\cdot 0$$
$$\dots$$
