Prove that $\prod_{i=1}^n (1+x_i/n) \sim \exp (\sum_{i=1}^n x_i/n)$ as $n\rightarrow\infty$?

(I). $\lim_{n\to \infty}n^{-1}\sum_{j=1}^n x_j$ need not exist, even if $(x_j)_j$ is a bounded sequence.

(2). Let $A_n=\prod_{j=1}^n(1+x_j/n) .$ Let $B_n=\exp (n^{-1}\sum_{j=1}^n x_j).$ Let $n>C.$

We have $\forall j\leq n\,( |x_j|<n)$ so $$\log A_n=\sum_{j=1}^n\log (1+x_j/n)=$$ $$=\sum_{j=1}^n\sum_{s=1}^{\infty }(-1)^{s-1}(x_j/n)^s/s=$$ $$=(\sum_{j=1}^nx_j/n) + R_n$$ $$\text { where } \quad |R_n|=|\sum_{j=1}^n\sum_{s=2}^{\infty}(-1)^{s-1}(x_j/n)^s/s|\leq$$ $$\leq \sum_{j=1}^n \sum_{s=2}^{\infty}(C/n)^s/s=n\sum_{s=2}^{\infty}(C/n)^s/s=$$ $$=(C^2/n)\sum_{t=0}^{\infty} (C/n)^t/(t+2)\leq$$ $$\leq (C^2/n)\sum_{t=0}^{\infty}(C/n)^t=(C^2/n)\cdot 1/(1-C/n).$$ So $\lim_{n\to \infty}R_n=0.$

But we also have $R_n=\log A_n-\log B_n=\log(A_n/B_n).$ So $\lim_{n\to \infty}\log (A_n/B_n)=0.$ So $\lim_{n\to \infty}(A_n/B_n)=1.$

So if the Cesaro mean $\lim_{n\to \infty}n^{-1}\sum_{j=1}^n x_j$ exists then the LHS and RHS in your Q are equal. If the Cesaro mean does not exist, then the ratio of the LHS to the RHS still converges to $1.$

Given $$ S(n) = \prod\limits_{i = 1}^n {\left( {1 + {{x_{\,i} } \over n}} \right)} \quad \left| {\;\left| {x_{\,i} } \right| \le C \le \left\lceil C \right\rceil = D} \right. $$ let's take $D<<n$ so that we can put $n=mD\quad | \; 2 \le m \in \mathbb Z$.
Thereafter taking the logarithm we get $$ \eqalign{ & \ln S(n) = \sum\limits_{i = 1}^n {\ln \left( {1 + {{x_{\,i} } \over n}} \right)} = \sum\limits_{i = 1}^{mD} {\ln \left( {1 + {{x_{\,i} } \over {mD}}} \right)} = \cr & = \sum\limits_{i = 1}^{mD} {\ln \left( {1 + {{\left( {x_{\,i} /D} \right)} \over m}} \right)} \cr} $$

The Lagrange formulation of the Taylor remainder gives $$ \ln \left( {1 + z} \right)\quad \left| {\,\left| z \right|} \right. < 1/2 = z - {1 \over {2\left( {1 + \zeta } \right)^{\,2} }}z^{\,2} \quad \left| {\, - 1/2 < \zeta } \right. < 1/2 $$ so that $$ \left| {\, - {1 \over {2\left( {1 + \zeta } \right)^{\,2} }}\;} \right|z^{\,2} < 2z^{\,2} $$

Therefore we can write $$ \eqalign{ & \ln S(n) = \sum\limits_{i = 1}^{mD} {\ln \left( {1 + {{\left( {x_{\,i} /D} \right)} \over m}} \right)} = \sum\limits_{i = 1}^{mD} {\left( {{{\left( {x_{\,i} /D} \right)} \over m} + O\left( {{1 \over {m^{\,2} }}} \right)} \right)} = \cr & = \sum\limits_{i = 1}^{mD} {{{\left( {x_{\,i} /D} \right)} \over m} + O\left( {{1 \over m}} \right)} = \sum\limits_{i = 1}^n {{{x_{\,i} } \over n} + O\left( {{1 \over {n/D}}} \right)} \cr} $$ and $$ \eqalign{ & \mathop {\lim }\limits_{n\, \to \;\infty } \ln S(n) = \mathop {\lim }\limits_{n\, \to \;\infty } \left( {\sum\limits_{i = 1}^n {{{x_{\,i} } \over n} + O\left( {{1 \over {n/D}}} \right)} } \right) = \cr & = \mathop {\lim }\limits_{n\, \to \;\infty } \left( {{1 \over n}\sum\limits_{i = 1}^n {x_{\,i} } } \right) \cr} $$

Thus, whether the two limits exist or not, their ratio is in any case $1$.