Note that, for the least significant bit, addition mod $2^n$ and xor are the same. So if the cipher doesn't include rotation then you can set up a linear equation in the least significant bits of the plaintext, ciphertext, and round keys that holds with probability 1 (if the cipher breaks the state up into two or more distinct parts, like a Feistel cipher, then you need to consider the least significant bits of each part). That means with a single known plaintext you can extract a single bit of information about the least significant bits of the round keys.
This single bit of information narrows down the possibilities for the least significant carry bits, which enables you to gain one more bit of information about the LSBs of the round keys by observing what happens to the least 2 significant bits of the plaintext and ciphertext. That narrows down the possibilities for the LSB carry bits even further, enabling you to gain another bit by looking at the least 3 significant bits of the plaintext and ciphertext, and so on. By progressively narrowing down the possibilities for the round key LSBs, you can gain complete knowledge of those bits (so long as the number of rounds does not exceed the 'part' size of the cipher), and then you can attack the 2nd LSBs using the same techniques.
To illustrate, take this very simple example cipher: $E(x) = ((((x + R_0) \oplus R_1) + R_2) \oplus R_3)$, where the cipher operates on blocks of $n$ bits, "+" is addition over $2^n$, $\oplus$ is xor, and each 'round' key $R_i$ is an $n$-bit independently selected random string. From this, if $X^j$ denotes the $j$th bit of a given word (zero being the least significant bit) and $P$ and $C$ denote the plaintext and ciphertext respectively, given a single known plaintext the following equation gives you a single bit of information about the four round keys:
$$P^0 \oplus C^0 = R_0^0 \oplus R_1^0 \oplus R_2^0 \oplus R_3^0.$$
Let us suppose that the result is zero. That means either all four least significant bits of the round keys are zero, or two of them are zero, or none of them are zero - there are eight possible arrangements in total. If you get two known/chosen plaintexts where $P^1$ is fixed to some value and $P^0$ varies, and look at the resulting $C^0$ and $C^1$, you can narrow those 8 possibilities for the LSBs of the round keys down to 4 possibilities. In a similar fashion those four can then be narrowed down to two, and then to one, using a few more known/chosen plaintexts.
With the influence of the least significant carry bits known, you can apply the same technique to determine the four $R^1$ bits. That way you can progressively reveal almost all the bits of the round keys, using relatively few known/chosen plaintexts.
Because of this sort of vulnerability in using only addition and xor, ARX ciphers include rotation (that's the 'R' part of the name).