Summarizing the question:
Would ECDSA-256 still provide 128 bit security for a 128 bit private key padded to 256-bit?
No, for fixed public 128-bit padding. Given ECDSA curve parameters, the ECDSA public key $Q$ and the padding method that produced the private key $d$, it's possible to devise an attack that finds the private key $d$ with $Q=dG$ using about $2^{65}$ point additions, that is like $65$-bit security.
Left padding extends a 128-bit secret $s$ to $d=k\mathbin\| s=2^{128}k+s$ for some known 128-bit $k$. Thus the problem is to find $s$ given $Q=(2^{128}k+s)G$, that is find $s$ such that $sG=Q-2^{128}kG$. The right hand side can be readily computed. That $s$ can be found using Baby Step/Giant Step, or Pollard's rho.
For right padding, $d=s\mathbin\|k=2^{128}s+k$ and the problem is to find $s$ given $Q=(2^{128}s+k)G$, that is find $s$ such that $s(2^{128}G)=Q-kG$, which is equally easy.
On the other hand, if we build $d$ from $s$ using a hash, for example as $d=(\operatorname{SHA-512}(s)\bmod(n-1))+1$, then we get 128-bit security for single target attack (that is when the adversary attacks a single public key $Q$).
In multi-target attack, the attacker has a collection of $r$ public keys $Q_i$ and is content with finding any $d$ with $dG$ among the $Q_i$. Even with 128-bit to 256-bit expansion with a hash, an attack that simply tries various $s$ (e.g. sequentially) succeeds with about $2^{128}/r$ hashes and scalar multiplications, thus security can't exceed like $\min(136-\log_2(r),128)$ bit.
If we want multi-target security with a 128-bit secret and no diversifier/salt, we need some level of key stretching with e.g. Argon2.
