I only get

$$\sum_{p\le y}\frac{1}{p^{1-\delta}}\le \frac{y^{\delta}}{\log(y^\delta)}
+e^2\log(1/\delta)+O\Bigl(\frac{y^\delta}{\delta^2(\log y)^2}+1\Bigr).$$

The first sum is
$$\sum_{p\le e^{2/\delta}}\frac{1}{p^{1-\delta}}=
\sum_{p\le e^{2/\delta}}\frac{p^\delta}{p}\le e^2
\sum_{p\le e^{2/\delta}}\frac{1}{p}$$
and applying the known bound on the sum of reciprocal of primes we get
$$e^2\log\log e^{2/\delta}+ C+O{(\log x)^{-1}}= e^2\log(1/\delta)+O(1).$$

For the second sum for $2^{1/\delta}<p\le y$ we have

$$=
\frac{\pi(y)}{y^{1-\delta}}-
\frac{\pi(2^{1/\delta})}{2^{(1-\delta)/\delta}}+(1-\delta)\int_{2^{1/\delta}}^y\pi(t)
t^{-2+\delta}\,dt\le$$
$$\le \frac{y^\delta}{y}\Bigl(\frac{y}{\log y}+O(y/(\log y)^2)\Bigr)+
(1-\delta)\int_{2^{1/\delta}}^y\frac{t}{\log t}
t^{-2+\delta}\,dt+$$
$$O\Bigl(\int_{2^{1/\delta}}^y\frac{t}{(\log t)^2}
t^{-2+\delta}\,dt\Bigr)=$$
$$\le\frac{y^{\delta}}{\log y}+O\Bigl(\frac{y^{\delta}}{(\log y)^2}\Bigr)+
\frac{1-\delta}{\delta}\frac{y^\delta}{\log y}+
O\Bigl(\frac{y^\delta}{\delta^2(\log y)^2}\Bigr).$$