= Solution
Put $A=\Sigma^\infty X_+$ and $H=F(A,E)=F(X_+,E)$. The diagonal of $X$ and its map to a point induce
$$
\Delta:A\longrightarrow A\wedge A,\qquad\epsilon:A\longrightarrow\mathbb S.
$$
Here the identification $(X\times X)_+\cong X_+\wedge X_+$ explains the target of $\Delta$. These maps satisfy the coassociativity, cocommutativity and counit identities, since their underlying maps send $x$ to repeated copies of the same $x$.
Use the defining adjunction of the <function spectrum> to define $m:H\wedge H\to H$. Its adjoint is the composite
$$
H\wedge H\wedge A\xrightarrow{1\wedge\Delta}H\wedge H\wedge A\wedge A\xrightarrow{\text{interchange}}(H\wedge A)\wedge(H\wedge A)\xrightarrow{\mathrm{ev}\wedge\mathrm{ev}}E\wedge E\xrightarrow{\mu}E.
$$
Define $u:\mathbb S\to H$ to have adjoint
$$
A\xrightarrow{\epsilon}\mathbb S\xrightarrow{\eta}E.
$$
Thus $m$ is pointwise multiplication, while $u$ is the constant unit. These are genuine stable maps, defined through the internal <function spectrum>, so the argument does not presume that stable functions have ordinary points.
To verify associativity, take the adjoints of $m(m\wedge1)$ and $m(1\wedge m)$. Both are maps $H^{\wedge3}\wedge A\to E$. Both use the triple diagonal $A\to A^{\wedge3}$, evaluate the three copies of $H$, and multiply the three outputs. The triple diagonals agree by coassociativity; the output maps agree by associativity of the <ring spectrum> $E$. The <function spectrum> adjunction therefore makes the original two maps equal. For the left unit, the adjunct of $m(u\wedge1)$ uses $(\epsilon\wedge1)\Delta=1_A$ and $\mu(\eta\wedge1)=1_E$, leaving exactly evaluation. Its adjoint is $1_H$. The same calculation with $(1\wedge\epsilon)\Delta$ proves the right unit. Hence
$$
\boxed{F(X_+,E)\text{ is a ring spectrum}.}
$$
If $E$ is a <commutative ring spectrum>, cocommutativity of $\Delta$ proves that this <function spectrum> is commutative as well.
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