Let $X \subseteq \mathbf{P}^4$ be a hypersurface over $k$ algebraically closed. Why do we have $Pic(X) = Pic(\mathbf{P}^4)$ by the Lefschetz hypersurface theorem? I only see for $\mu_n$ coefficients $H^1_{et}(X,\mu_n) = H^1_{et}(\mathbf{P}^4,\mu_n)$ and $H^1_{et}(X,\mu_n) = Pic(X)[n]$, so the torsion subgroups are identified. Perhaps $NS(X) = \mathbf{Z}$?
OK, we also have $0 \to Pic(X)/n \to H^2(X,\mu_n) \to Br(X)[n] \to 0$ and the $H^2$ coincide, so $Pic(-) \otimes \mathbf{Z}_\ell$ coincide.
I am not aware of a proof along the lines you suggest: Grothendieck in SGA 2, Exposé XII, Corollaire 3.7 gives a proof based on formal schemes.
However, one should take into account that SGA II precedes the SGA volumes on étale cohomology so that in a parallel universe where EGA is completed (and includes all of SGA, just as intended by Grothendieck) there might be an étale-cohomological proof of Grothendieck-Lefschetz.
Over $\mathbb C$ there is a proof of the theorem you mention in the spirit of your suggestion.
It is based on the long exact sequence derived from the exponential exact sequence $$0\to \underline {\mathbb Z}\to \mathcal O_Y\to \mathcal O_Y^\ast\to 0$$ You have to replace successively $Y$ by $\mathbb P^n$ and $X$ and compare the two long exact sequences.
Remember that $H^1(X,\mathcal O_X)=H^2(X,\mathcal O_X)=0$, which follows from the exact sequence $$ 0\to O_{\mathbb P^4}(-d)\to \mathcal O_{\mathbb P^4}\to \mathcal O_X\to 0 $$(where $d$ is the degree of $X$), again by taking cohomology.
Remark
Some readers might ask why we have this strange hypothesis on the dimension of projective space.
The reason is that the the theorem is completely false in $\mathbb P^1$, $\mathbb P^2$ and $\mathbb P^3$: a curve in $\mathbb P^2$ has in general a huge Picard group (actually an algebraic variety) and a smooth quadric in $\mathbb P^3$ already has Picard group $\mathbb Z\times \mathbb Z$. (Need I recall that $\text {Pic}(\mathbb P^m)=\mathbb Z$ ?)
I'll let these readers ponder on hypersurfaces of $\mathbb P^1 \dots$