Indeed, ECB is such that encrypting twice the same plaintext leads to the same ciphertext. Even worse, encrypting a plaintext containing twice the same plaintext block leads to a ciphertext containing twice the same ciphertext block. Either is a disadvantage because it goes against the ideal of a cipher: depriving the adversary from any knowledge about the plaintexts, with the sole exception of length.
An example of why that's a drawback is if we encipher plaintexts consisting of votes cast in an election, with plaintext N for a no vote, Y for a yes vote. With any cipher such that encrypting twice the same plaintext leads to the same ciphertext, someone intercepting the ciphertexts will see two values, and can determine each vote (assuming a single vote is known, or it is known that Y outnumbers N).
If a cipher has the property that plaintext and ciphertext have the same size for any unconstrained plaintext, then indeed (in order for plaintext to be exactly decipherable, and by a counting argument) the cipher is bound to have said drawback. Hence modern ciphers have a ciphertext larger than the plaintext (typically for symmetric ciphers: by the Initialization Vector's size at least).
Addition per comment: an example of a cipher not bound to generate the same ciphertext when enciphering twice the same plaintext would be $\operatorname{AES}$ in counter mode with key $K$
- encryption:
- generate a 128-bit Initialization Vector $C$ at random, using the computer equivalent of tossing a coin 128 times, and outputs $C$ as the first 128 bits of the ciphertext
- while remaining plaintext length $l>0$
- output the eXclusive-OR of the next $\max(l,128)$ bits of plaintext and as many of the leftmost bits of $\operatorname{AES-ENC}_K(C)$
- $C\gets C+1\bmod2^{128}$ (increment the counter)
- decryption:
- set $C$ as the first 128 bits of the ciphertext
- while remaining ciphertext length $l>0$
- output the eXclusive-OR of the next $\max(l,128)$ bits of ciphertext and as many of the leftmost bits of $\operatorname{AES-ENC}_K(C)$
- $C\gets C+1\bmod2^{128}$ (increment the counter)
