If e1 is selected even, the algorithm strength remain unchanged?
Actually, as far as we know, this algorithm relies on a weaker security assumption (and hence is arguably stronger than RSA).
RSA is known to be breakable if you can factor the modulus; however it's possible that there is a way to break RSA without factoring. We don't know of such a way, but it may exist.
In contrast, it is well known that you can efficiently factor $n$ if you are given a method that takes $x^e \bmod n$, for $e$ even, and recovers a possible $x$ value. Hence there cannot be a (much) easier way to reconstruct $x$ than factoring $n$
On the other hand:
what is a different way to public encryption key and ensure everybody that I do not posses decryption key?
What makes you think you can convince anyone that, given $x^e \bmod n$ for even $e$, you cannot find $x$? If you know the factorization of $n$ (which, assuming that you are the guy that picked $n$, is a reasonable assumption), you can reconstruct a handful of possible $x$ values (no more than $\gcd(p-1, e) \times \gcd(q-1, e)$, which may be as small as 4), one of which is the correct value.
Now, what is the problem you're trying to solve? If you want to make sure that there is no 'decryption key', why don't you just, say, do a hash of the plaintext.
If there needs to be a 'decryption key' (and just that no one knows it), well, one way is to use one of the larger RSA challenge numbers as a normal RSA modulus (and use a large prime public exponent). However, we rarely need to do that; what are you trying to accomplish?