Addendum: I just realised that I didn't answer the actual question aboutwhether there is a simpler proof (to the isogeny statement would be myinterpretation). I doubt it (depending of course somewhat on your definition ofsimpler). A proof (arguably less simple) avoiding the use of Chevalley's theorem(but assuming to be on the safe side that the base field is algebraically closedof characteristic $0$) would be to prove that for some $n$ the pull back of theextension along the $n$'th power on $\mathbb{G}_m$ is trivial as an $A$-torsorbecause then the extension would be described by a $2$-cocycly$\mathbb{G}_m\times\mathbb{G}_m\rightarrow A$ and any map of that form isconstant. That in turn will follow from the fact that for some $m$ the class ofthis extension as an $A$-torsor would be in the image of$H^1(\mathbb{G}_m,A[m])\rightarrow H^1(\mathbb{G}_m,A)$. This follows from thefact that $H^1(\mathbb{G}_m,A)$ is torsion and then the conclusion follows asany class of $H^1(\mathbb{G}_m,C)$ for any finite locally constant sheaf $C$ iskilled by pulling back along some $n$'th power map.

Using the Kummer exact sequence $0\rightarrow\mu_n\rightarrow\mathbb{G}_m\rightarrow\mathbb{G}_m\rightarrow0$ we get a long exact sequence $$0\rightarrow\mathrm{Hom}(\mathbb{G}_m,A)\rightarrow\mathrm{Hom}(\mathbb{G}_m,A) \rightarrow\mathrm{Hom}(\mu_n,A)\rightarrow\mathrm{Ext}^1(\mathbb{G}_m,A).$$ As $\mathrm{Hom}(\mathbb{G}_m,A)=0$ this gives an embedding $\mathrm{Hom}(\mu_n,A)\hookrightarrow\mathrm{Ext}^1(\mathbb{G}_m,A)$ and at least over an algebraically closed field in which $n$ is invertible we always have that $\mathrm{Hom}(\mu_n,A)$ is non-trivial and consequently so is $\mathrm{Ext}^1(\mathbb{G}_m,A)$.
This does not contradict Deligne's claim as he seems to be (somewhat implicitly it must be admitted) speaking of extensions up to isogeny (in Principle 2.1 this is made more clear). In fact Chevalley's theorem implies that this is true: Consider a connected affine subgroup $G'\hookrightarrow G$ for which the quotient is an abelian variety. Then the kernel of the composite $G'\rightarrow\mathbb{G}_m$ is affine and embeds in $A$ and thus is finite. On the other hand $G'\rightarrow\mathbb{G}_m$ must be surjective as otherwise $G'$ would be finite and hence trivial so $G$ would be an abelian variety which is not possible. The kernel of $G'=\mathbb{G}_m\rightarrow\mathbb{G}_m$ is then some $\mu_n$ which shows that the extension is in the image of $\mathrm{Hom}(\mu_n,A)\hookrightarrow\mathrm{Ext}^1(\mathbb{G}_m,A)$ and in particular is trivial up to isogeny.
The final conclusion is that $\mathrm{Ext}^1(\mathbb{G}_m,A)=\mathrm{Hom}(\hat{\mathbb Z}(1),A)$.
Using the Kummer exact sequence $0\rightarrow\mu_n\rightarrow\mathbb{G}_m\rightarrow\mathbb{G}_m\rightarrow0$ we get a long exact sequence $$0\rightarrow\mathrm{Hom}(\mathbb{G}_m,A)\rightarrow\mathrm{Hom}(\mathbb{G}_m,A) \rightarrow\mathrm{Hom}(\mu_n,A)\rightarrow\mathrm{Ext}^1(\mathbb{G}_m,A).$$ As $\mathrm{Hom}(\mathbb{G}_m,A)=0$ this gives an embedding $\mathrm{Hom}(\mu_n,A)\hookrightarrow\mathrm{Ext}^1(\mathbb{G}_m,A)$ and at least over an algebraically closed field in which $n$ is invertible we always have that $\mathrm{Hom}(\mu_n,A)$ is non-trivial and consequently so is $\mathrm{Ext}^1(\mathbb{G}_m,A)$.
This does not contradict Deligne's claim as he seems to be (somewhat implicitly it must be admitted) speaking of extensions up to isogeny (in Principle 2.1 this is made more clear). In fact Chevalley's theorem implies that this is true: Consider a connected affine subgroup $G'\hookrightarrow G$ for which the quotient is an abelian variety. Then the kernel of the composite $G'\rightarrow\mathbb{G}_m$ is affine and embeds in $A$ and thus is finite. On the other hand $G'\rightarrow\mathbb{G}_m$ must be surjective as otherwise $G'$ would be finite and hence trivial so $G$ would be an abelian variety which is not possible. The kernel of $G'=\mathbb{G}_m\rightarrow\mathbb{G}_m$ is then some $\mu_n$ which shows that the extension is in the image of $\mathrm{Hom}(\mu_n,A)\hookrightarrow\mathrm{Ext}^1(\mathbb{G}_m,A)$ and in particular is trivial up to isogeny.