online-tools | Online tools provides md2 md5 sha1 sha2 sha512 | Cryptography library

You need to initialize the IV to all zeroes because otherwise it starts with a random value:

Convert the hexadecimal string to bytes using bytes.fromhex():

openssl requires Hex encoding of manual keys and IVs, not ASCII encoding. (ASCII or UTF-8 encoding these, as the online tools seems to be doing, dramatically reduces your keyspace and makes AES insecure.)

"E" is 0x45 and "F" is 0x46, so the equivalent openssl command is:

You must decode hex string first.

After getting corresponding raw value, you can convert it to Base64 string.

The website uses neither a padding nor an IV. Therefore in the code the padding must be disabled and the ECB mode must be applied.

Furthermore the website expects a hex encoded key and ciphertext and returns the decrypted data also hex encoded, which therefore must not be UTF-8 decoded in the code:

The devlan site explains

CBC mode is highly recommended and it requires IV to make each message unique. If no IV is entered then default will be used here for CBC mode and that defaults to a zero based byte[16].

So your prepended IV is considered as a ciphertext block. Now, you have two blocks there. In the CBC mode, the decryption is performed as ( blocks are counted form 1);

In the PHP code AES-128 is specified. Therefore PHP implicitly truncates the 32 bytes key to the first 16 bytes. In the CryptoJS code only this 16 bytes key may be used.
Furthermore, key and IV are converted most easily with the CryptoJS encoders into a WordArray.
A possible CryptoJS implementation is:

PKCS 5 (or 7) padding is not adding 0 bytes, but adding a c byteswith valuec(where1 <= c <= 16) if you're c` bytes short of a block length multiple.

So if you already have a multiple of 16, add a full 16 bytes of value 16, and if your last block is 'stop' (4 bytes), we add 12 bytes with value 0xc (12 in hex) to fill up the block. Etc.

This way the receiver (after decryption of the final block) can check the last byte c and check if the value is 1 <= c <= 16 (if not, reject the decryption) and then check that the last c bytes indeed are all that same value, and then remove them from the decryption. This way the receiver does not have to guess how many bytes of the last block were only padding or really part of the plain text. It's unambiguous doing it the PKCS way.

I'll leave the coding up to you.