I have now understood the situation better so my previous post has been replaced by this. (The only thing that was in the original but will not be here are some explicit formulas but Theo has given a reference for that.)
As I understand the question the poser wanted a construction of the enveloping algebra of a Lie algebra in a symmetric monoidal pseudoabelian (i.e., idempotents have kernels) $K$-category $\mathcal C$ with arbitrary sums over a field $K$ of characteristic zero. This means that for any $K[\Sigma_n]$-module $M$ and any object $V\in\mathcal C$ we can define $M\bigotimes_{\Sigma_n}V^{\otimes n}$ and for any $\Sigma$-module $M_\bullet$ (i.e., a collection $(M_n)$ of $K[\Sigma_n]$-modules) we can define $M(V):=\bigoplus_nM_n\bigotimes_{\Sigma_n}V^{\otimes n}$ which is an endofunctor of $\mathcal C$. Furthermore, a map $M \to N$ of $\Sigma$-modules gives a natural transformation of functors $M(V) \to N(V)$. In the particular case when $\mathcal C$ is the category of $K$-vector spaces such a natural transformation comes from a unique map of $\Sigma$-modules. The idea is to do what we know to do for $K$-vector spaces, interpret it as a set of natural transformations, get the corresponding maps of $\Sigma$-modules and use them to induce natural transformations for a general $\mathcal C$.
Let thus $S(V)$ be the symmetric algebra on the $K$-vector space $V$, $T(V)$ the tensor algebra on $V$ and $L(V)$ the free Lie algebra on $V$. Symmetrisation gives an isomorphism $S(L(V)) \to U(L(V))=T(V)$, where $U(-)$ is the enveloping algebra of Lie algebras. As $T(V)$ is a $T$-algebra (i.e., an associative algebra) and we can use this isomorphism to give $S(L(V))$ a $T$-algebra structure (i.e., a natural transformation $T(S(L(V))) \to S(L(V))$ fulfilling the appropriate conditions with respect to the monad structure on $T(-)$). Furthermore, if $\mathfrak g$ is a Lie algebra, then the $T$-algebra structure on $S(\mathfrak g)$ induced by the isomorphism $S(\mathfrak g) \to U(\mathfrak g)$ is given as the composite of $T(S(\mathfrak g)) \to T(S(L(\mathfrak g)))$ induced by the inclusion $\mathfrak g \to L(\mathfrak g)$, the map $T(S(L(\mathfrak g))) \to S(L(\mathfrak g))$ given by the $T$-module structure on $S(L(V))$ above and the map $S(L(\mathfrak g)) S(\mathfrak g)$ induced by the structure map $L(\mathfrak g) \to \mathfrak g$
Now, the functors $S(-)$, $L(-)$ and $T(-)$ are associated to $\Sigma$-modules which will be denoted by the same letters (instead of the standard $Com$, $Lie$ and $Ass$). Furthermore, composition of functors correspond to the plethysm $\circ$. Hence we get that $S\circ L$ is a $T$-module, i.e., we have a map $T\circ S\circ L \to S\circ L$ compatible with the operad structure on $T$. Consider now the case of a general $\mathcal C$. Each of $S$, $L$ and $T$ give endofunctors on $\mathcal C$ and $\circ$ again corresponds to composition. Let $\mathfrak g$ be a Lie algebra in $\mathcal C$ and define a $T$-algebra structure (i.e., the structure of associative algebra) on $S(\mathfrak g)$ as the composite $$ T(S(\mathfrak g)) \to T(S(L(\mathfrak g))) \to S(L(\mathfrak g)) \to S(\mathfrak g) $$ as above. The verification that this does indeed give a $T$-algebra structure is just a question of unwinding the definitions. The fact that $S$ is an operad gives us a natural transformation $V \to S(V)$ which applied to $\mathfrak g$ gives a morphism $\mathfrak g \to S(\mathfrak g)$ which we now want to show is a Lie algebra homomorphism. Here the Lie algebra structure on $S(\mathfrak g)$ is induced by its $T$-algebra structure and the operad map $L \to T$. Again unwinding definitions shows that it is indeed a Lie algebra morphism.
Finally assuming that $\mathfrak g \to A$ is a Lie algebra homomorphism where $A$ is an associative algebra with $L \to T$ inducing its Lie algebra structure. Note that we have an isomorphism (now going back to vector spaces) $S(L(V))\to T(V)$ and hence an isomorphism is $\Sigma$-modules $S\circ L=T$. This gives us a map $S(\mathfrak g)\to S(L(\mathfrak g))=T(\mathfrak g) \to T(A) \to A$ and it is easy to see that this is an algebra morphism.
