Just by looking at the answers and comments you have made I can see that your frustration with multiplication is a deep one. Most of the answers revolve precisely around the fact that it just works! But that's not enough.
Your question reminds me a lot of a similar question. Suppose we have a rectangle with sides $a$ and $b$. Then its area is given by $a\cdot b$ and not by $a+b$. The question might very well be, why do we use multiplication and not addition.
Here is my slightly different take on this question.
A good question might be what is multiplication really? The obvious answer is, it is derived from repeated addition. It's much neater to write $5\cdot 7$ instead of $5+5+5+5+5+5+5$. In the exact same way it's much neater to write $3^5$ to mean $3\cdot3\cdot3\cdot3\cdot3$. You get a whole sequence of mathematical operations starting from addition. Then you have multiplication which is repeated addition. Then you have exponentiation which is repeated multiplication. Next you have tetration. These are shorthands for making things easier to write out.
This still does not answer the question "why multiplication". I can answer why it's not addition. It's because addition is used in a different situation.
Suppose you have a set of 20 kids - a set $\{b_1,\dots,b_7\}$ of 7 boys, and a set $\{g_1,\dots,g_{13}\}$ of 13 girls. You can ask different questions about these sets. For example - "How many kids are there in the class?". You can answer this in the simple way 7+13 = 20. Can we do something with the sets themselves to get the same answer? We'll have to combine them with a "set operation". It's called "disjoint union". This is just putting the two sets next to each other and making a new set: $\{g_1,\dots,g_{13},b_1,\dots,b_7\}$. Taking the size of this set gives us 20.
Now suppose that they are all taking a dance class together. A question we can ask is this "How many different couples can we make out of the two sets of kids?". The answer is $7\cdot 13 = 91$ different couples. This corresponds to a different set operation. You are taking the two sets and combining them in a different way - a set "product". Each girl can dance with each boy, so to each girl corresponds a whole set of boys. Putting the sets together gives us something like $\{(g_1,b_1),\dots,(g_1,b_7),\dots,(g_{13},b_7)\}$. A good way to think about this is putting the sets perpendicular to each other and making a table, where each entry is the given pair.
$\begin{bmatrix}(g_1,b_1)& (g_1,b_2)& \cdots & (g_1,b_7)\\ (g_2,b_1)& (g_2,b_2)& \cdots & (g_2,b_7) \\ \vdots& \vdots& \ddots & \vdots\\ (g_{13},b_1)& (g_{13},b_2)& \cdots & (g_{13},b_7)\end{bmatrix}$
You may complain that my reasoning is circular and it is. It's just a different angle on the same problem.