Instead of using exp, I'll show how to derive $\int_0^t f(B_t)\,dB_t$, from the definition, assuming $\,f''$ exists and is continuous.
Let $F(x)=\int_0^xf(s)\,ds$. We of course start with a partition $0=t_0<t_1<\dots<t_m=t$, and setting $B_k=B_{t_k}$. Let $\mu=\sup_k t_{k+1}-t_k$ be the mesh of the partition.
Write
$$
F(B_{k+1})=F(B_k)+f(B_k)(B_{k+1}-B_k)+\frac12f'(B_k)(B_{k+1}-B_k)^2+R_k(B_{k+1})
$$
where $R_k(x)$ is the remainder term for the Taylor series centered at $B_k$. Rearranging, and letting $\Delta B_k=B_{k+1}-B_k$,
$$
f(B_k)\Delta B_k=F(B_{k+1})-F(B_{k})-\frac12 f'(B_k)(\Delta B_k)^2-R_k(B_{k+1})
$$
We then sum the above expression over $0\le k\le m-1$, and take the limit in probability as $\mu\to 0$ (meaning we are really using a sequence of partitions). The LHS of above becomes $\int f(B_t)dB_t$, by definition.
On the right, the $F(B_{k+1})-F(B_{k})$ part is telescoping, adding up to $F(t_n)-F(t_0)=F(t)$.
To evaluate the $f'(B_k)(\Delta B_k)^2$ part, we let $\Delta t_k=t_{k+1}-t_k$, and replace this with
$$
f'(B_k)\Delta t_k + f'(B_k)((\Delta B_k)^2 - \Delta t_k)
$$
Then $\sum_{k=0}^{m-1}f'(B_k)\Delta t_k \to \int_0^t f'(B_k)dt$, and $\sum_{k=0}^{m-1}f'(B_k)(\Delta B_k - \Delta t_k)\to 0$. You show the last limit approaches zero by computing its variance:
$$
Var\left(\sum_{k=0}^{m-1}f'(B_k)((\Delta B_k)^2 - \Delta t_k)\right)=\sum_{k=0}^{m-1}Var(f'(B_{k}))(\Delta t_k)^2 Var\left(\frac{\Delta (B_k)^2}{\Delta t_k}-1\right)$$
$$\qquad \qquad \qquad\le Ct\cdot \mu\cdot \kappa\cdot \sum \Delta t_k=Ct\mu \kappa t\stackrel{\mu\to 0}\to 0$$
where $C=\sup_{s\in [0,t]} f'(s)$, so that $Var(f'(B_k))\le Var(CB_k)\le C^2 t_k\le C^2t$, and $\kappa$ is the variance of a $\chi^2$ random variable (note that $\frac{\Delta (B_k)^2}{\Delta t_k}$ are i.i.d. $\chi^2$). Since the variance $\to0$, and its mean is zero, the LHS $\to 0$ in $L_2$.
Finally, letting $K=\sup_{0\le s\le t}f''(t)$, the remainder term $R(B_{k+1})$ is uniformly bounded by $K(\Delta B_k)^3$, and a similar variance calculation shows that $\sum_{k=0}^{m-1} R_k(B_{k+1})\to 0$.
Putting this all together,
$$
\int_0^t f(B_k)\,d B_t=F(B_t)-\frac12\int_0^t f'(B_k)\,dt
$$
There's actually two definitions of solution to SDE. Strong and weak.
Strong solution. Given a probability space $
(\Omega, \mathcal F, \mathcal F_t,P)$ and a Brownian motion $B(t)$ on that space adapted to $\mathcal F_t$, a solution to $dX=fdt+g~dB(t)$ with initial condition $x\in \mathbb R$ is a continuous process adapted to $\mathcal F_t$ such that $X(t)=x+\int_0^t fdt+\int_0^t g~dB(t)$.
There is also
Weak solution. A process $X(t)$ is called a weak solution to $dX=fdt+g~dB(t)$ if there exists a probability space $
(\Omega, \mathcal F, \mathcal F_t,P)$ and a Brownian motion $B(t)$ on that space adapted to $\mathcal F_t$ such that $X(t)=x+\int_0^t fdt+\int_0^t g~dB(t)$.
Note that strong solution implies weak solution.
As an example of something that has weak but not strong solution, consider Tanaka's equation
$$dX(t)=\text{sign}(X(t))dB(t)$$
It can be shown that this has no strong solution by considering local times. However let $X(t)$ be a Brownian motion and then $X(t)$ is a weak solution. Note that $\int_0^t\text{sign}^2(s)ds=t<\infty$ so the Ito integral $\int_0^t\frac{1}{\text{sign}(s)}dX(s)=\int_0^t \text{sign}(s) dX(s)$ exists. Noting that the quadratic variation of this integral is $t$, and Ito integrals are continuous martingales, so we know $\int_0^t\frac{1}{\text{sign}(s)}dX(s)=\tilde{B}(t)$ is a Brownian motion.
Thus, $dX(t)=\text{sign}(t)d\tilde{B}(t)$.
The difference here is that the Brownian motion is constructed in terms of our solution not the other way around.
Best Answer
The solution of an Itö-SDE $X_{t}=y+\int bdt+\int \sigma dB_{t}$ is defined outside a set of measure zero that depends on the whole equation i.e. $\Omega_{y,b,\sigma}$ with $P[\Omega_{y,b,\sigma}]=0$.
Considering another initial condition $y$ will create a new set of measure zero outside of which the solution is defined. Since the set of allowed initial conditions is not countable, it is a priori not clear whether
$$P[\bigcup_{y}\Omega_{y,b,\sigma}]=0,$$
i.e. that there exists a set of full measure on which an Itö-SDE can be solved for every initial condition $y$. For a pathwise solution theory such as rough paths, this is immediate. In rough paths, the solutions are built deterministically by postulating the existence of some iterated integrals.
see here too On the pathwise uniqueness of solutions of SDEs(Stochastic Differential Equations).