The infinite symmetric product of a pointed metric space $(X,e)$ is metrizable iff the basepoint $e$ is isolated. First, if $e$ is isolated and $X=Y\coprod \{e\}$, then $SP(X,e)=\coprod SP^n(Y)$ and each finite symmetric product $SP^n(Y)$ is metrizable (by the metric you define). In fact, in this case it is not hard to see that your metric induces the topology on $SP(X,e)$.

Conversely, suppose $e$ is not isolated; then I claim $SP(X,e)$ is not first countable and hence not metrizable. Indeed, suppose $SP(X,e)$ is first countable. Since $e$ is not isolated, $[e]=[e,e,\dots]\in SP(X,e)$ is in the closure of $SP^n(X,e)\setminus SP^{n-1}(X,e)$ for each $n>0$. By first countability, we can find elements $x_n\in SP^n(X,e)\setminus SP^{n-1}(X,e)$ such that the sequence $(x_n)$ converges to $[e]$ in $SP(X,e)$. But the set $\{x_n\}$ has finite and hence closed intersection with each $SP^n(X,e)$, so the entire set $\{x_n\}$ is closed in $SP(X,e)$. This contradicts the assumption that $(x_n)$ converges to $[e]$.

As a final note, let $SP_d(X,e)$ be the symmetric product with the topology induced by your metric. Then the identity $i:SP(X,e)\to SP_d(X,e)$ is continuous, and in fact is a homeomorphism when restricted to each $SP^n(X,e)$. Since every compact subset of $SP(X,e)$ is contained in some $SP^n(X,e)$ (by essentially the same argument as the previous paragraph), the map $i$ is a weak equivalence (i.e., induces an isomorphism on homotopy groups). In particular, when $X$ is a metric space that is homotopy equivalent to a connected CW-complex (relative to the basepoint), the Dold-Thom theorem implies that $\pi_n(SP_d(X,e))=\pi_n(SP(X,e))=H_n(X,e)$.

The infinite symmetric product of a pointed metric space $(X,e)$ is metrizable iff the basepoint $e$ is isolated. First, if $e$ is isolated and $X=Y\coprod \{e\}$, then $SP(X,e)=\coprod SP^n(Y)$ and each finite symmetric product $SP^n(Y)$ is metrizable (by the metric you define). In fact, in this case it is not hard to see that your metric induces the topology on $SP(X,e)$.

Conversely, suppose $e$ is not isolated; then I claim $SP(X,e)$ is not first countable and hence not metrizable. Indeed, suppose $SP(X,e)$ is first countable. Since $e$ is not isolated, $[e]=[e,e,\dots]\in SP(X,e)$ is in the closure of $SP^n(X,e)\setminus SP^{n-1}(X,e)$ for each $n>0$. By first countability, we can find elements $x_n\in SP^n(X,e)\setminus SP^{n-1}(X,e)$ such that the sequence $(x_n)$ converges to $[e]$ in $SP(X,e)$. But the set $\{x_n\}$ has finite and hence closed intersection with each $SP^n(X,e)$, so the entire set $\{x_n\}$ is closed in $SP(X,e)$. This contradicts the assumption that $(x_n)$ converges to $[e]$. More generally, this argument shows that if a $T_1$ space $A$ is a colimit of subspaces $A_0\subset A_1\subset \dots$ such that the interiors of the $A_n$ do not cover $A$, then $A$ cannot be first countable.

As a final note, let $SP_d(X,e)$ be the symmetric product with the topology induced by your metric. Then the identity $i:SP(X,e)\to SP_d(X,e)$ is continuous, and in fact is a homeomorphism when restricted to each $SP^n(X,e)$. Since every compact subset of $SP(X,e)$ is contained in some $SP^n(X,e)$ (by essentially the same argument as the previous paragraph), the map $i$ is a weak equivalence (i.e., induces an isomorphism on homotopy groups). In particular, when $(X,e)$ is a pointed metric space that is homotopy equivalent to a connected CW-complex (relative to the basepoint), the Dold-Thom theorem implies that $\pi_n(SP_d(X,e))=\pi_n(SP(X,e))=H_n(X,e)$.

The infinite symmetric product of a pointed metric space $(X,e)$ is metrizable iff the basepoint $e$ is isolated. First, if $e$ is isolated and $X=Y\coprod \{e\}$, then $SP(X,e)=\coprod SP^n(Y)$ and each finite symmetric product $SP^n(Y)$ is metrizable (by the metric you define). In fact, in this case it is not hard to see that your metric induces the topology on $SP(X,e)$.

Conversely, suppose $e$ is not isolated; then I claim $SP(X,e)$ is not first countable and hence not metrizable. Indeed, suppose $SP(X,e)$ is first countable. Since $e$ is not isolated, $[e]=[e,e,\dots]\in SP(X,e)$ is in the closure of $SP^k(X,e)\setminus SP^{k-1}(X,e)$ for each $k>0$. By first countability, we can find elements $x^k=[x^k_n]\in SP^k(X,e)\setminus SP^{k-1}(X,e)$ such that the sequence $(x^k)$ converges to $[e]$ in $SP(X,e)$. WLOG, we may assume that for each $k$, the $x^k_n$ are ordered such that $d(x^k_n,e)$ is a nonincreasing function of $n$. Now let $U$ be the set of all $y=[y_n]\in SP(X,e)$ such that if the $y_n$ are ordered such that $d(y_n,e)$ is nonincreasing, $d(y_n,e)<d(x^n_n,e)$ for all $n$. This is clearly an open neighborhood of $[e]$ when restricted to each $SP^k(X,e)$, and hence it is an open neighborhood of $[e]$ in $SP(X,e)$. But no $x^k$ can be in $U$, contradicting the assumption that $(x^k)$ converges to $[e]$.

The infinite symmetric product of a pointed metric space $(X,e)$ is metrizable iff the basepoint $e$ is isolated. First, if $e$ is isolated and $X=Y\coprod \{e\}$, then $SP(X,e)=\coprod SP^n(Y)$ and each finite symmetric product $SP^n(Y)$ is metrizable (by the metric you define). In fact, in this case it is not hard to see that your metric induces the topology on $SP(X,e)$.

Conversely, suppose $e$ is not isolated; then I claim $SP(X,e)$ is not first countable and hence not metrizable. Indeed, suppose $SP(X,e)$ is first countable. Since $e$ is not isolated, $[e]=[e,e,\dots]\in SP(X,e)$ is in the closure of $SP^n(X,e)\setminus SP^{n-1}(X,e)$ for each $n>0$. By first countability, we can find elements $x_n\in SP^n(X,e)\setminus SP^{n-1}(X,e)$ such that the sequence $(x_n)$ converges to $[e]$ in $SP(X,e)$. But the set $\{x_n\}$ has finite and hence closed intersection with each $SP^n(X,e)$, so the entire set $\{x_n\}$ is closed in $SP(X,e)$. This contradicts the assumption that $(x_n)$ converges to $[e]$.

As a final note, let $SP_d(X,e)$ be the symmetric product with the topology induced by your metric. Then the identity $i:SP(X,e)\to SP_d(X,e)$ is continuous, and in fact is a homeomorphism when restricted to each $SP^n(X,e)$. Since every compact subset of $SP(X,e)$ is contained in some $SP^n(X,e)$ (by essentially the same argument as the previous paragraph), the map $i$ is a weak equivalence (i.e., induces an isomorphism on homotopy groups). In particular, when $X$ is a metric space that is homotopy equivalent to a connected CW-complex (relative to the basepoint), the Dold-Thom theorem implies that $\pi_n(SP_d(X,e))=\pi_n(SP(X,e))=H_n(X,e)$.