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.