What about using AES-ECB with random IV? Let P1P2...Pn be n blocks plaintext, and M0 be a random IV. Then a n+1 blocks message M0M1...Mn can be constructed by setting Mi+1 = Mi xor Pi+1. Finally, we use AES-ECB to encrypt this message. Is it as secure as using AES-CBC?
No, this is not a secure way of encrypting.  Specifically, it does not meet the requirements for indistinguishability under chosen plaintext attack (IND-CPA), a basic security definition for encryption.

According to IND-CPA, no attacker should be able to win the following game:


The attacker selects two equal-length plaintext messages.
The defender picks one of those plaintexts at random
The defender encrypts that plaintext and provides the ciphertext to the attacker
The attacker wins if it can determine which of the plaintexts was used


Here's a winning strategy for an attacker facing your scheme:


Let plaintext message 1 consist of a single block containing only zero bits.
Let plaintext message 2 be any other single-block message.
If the defender returns a ciphertext where the two ciphertext blocks are equal, then the attacker knows that plaintext message 1 was selected; else, the attacker knows that plaintext message 2 was selected.


Here's why this works:

In plaintext message 1, $P_1 = 0^n$ (i.e. $n$ zero bits), where $n$ is the number of bits in a block.
This means that $M_1 = M_0 \oplus P_1 = M_0 \oplus 0^n = M_0$,
so $M_1 = M_0$.
ECB mode produces ciphertext blocks $C_i = E_k(M_i)$, and $E_k$ is deterministic, so the ciphertext blocks will also match ($C_1 = C_0$).

In plaintext message 2, this equality will not hold: $P_1$ has at least one non-zero bit, so $M_1$ and $M_0$ will differ in at least one bit, resulting in different ciphertext blocks ($C_1 \neq C_0$).